YES(O(1), O(n^2)) 5.76/1.92 YES(O(1), O(n^2)) 5.76/1.96 5.76/1.96 5.76/1.96 5.76/1.96 5.76/1.96 5.76/1.96 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 5.76/1.96 5.76/1.96 5.76/1.96
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
5.76/1.96 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
5.76/1.96 
↳2 CdtProblem
5.76/1.96 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
5.76/1.96 
↳4 CdtProblem
5.76/1.96 
↳5 CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID))
5.76/1.96 
↳6 CdtProblem
5.76/1.96 
↳7 CdtLeafRemovalProof (ComplexityIfPolyImplication)
5.76/1.96 
↳8 CdtProblem
5.76/1.96 
↳9 CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
5.76/1.96 
↳10 CdtProblem
5.76/1.96 
↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
5.76/1.96 
↳12 CdtProblem
5.76/1.96 
↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
5.76/1.96 
↳14 CdtProblem
5.76/1.96 
↳15 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
5.76/1.96 
↳16 CdtProblem
5.76/1.96 
↳17 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
5.76/1.96 
↳18 CdtProblem
5.76/1.96 
↳19 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
5.76/1.96 
↳20 CdtProblem
5.76/1.96 
↳21 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
5.76/1.96 
↳22 CdtProblem
5.76/1.96 
↳23 SIsEmptyProof (BOTH BOUNDS(ID, ID))
5.76/1.96 
↳24 BOUNDS(O(1), O(1))
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5.76/1.96

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

if(true, t, e) → t 5.76/1.96
if(false, t, e) → e 5.76/1.96
member(x, nil) → false 5.76/1.96
member(x, cons(y, ys)) → if(eq(x, y), true, member(x, ys)) 5.76/1.96
eq(nil, nil) → true 5.76/1.96
eq(O(x), 0(y)) → eq(x, y) 5.76/1.96
eq(0(x), 1(y)) → false 5.76/1.96
eq(1(x), 0(y)) → false 5.76/1.96
eq(1(x), 1(y)) → eq(x, y) 5.76/1.96
negate(0(x)) → 1(x) 5.76/1.96
negate(1(x)) → 0(x) 5.76/1.96
choice(cons(x, xs)) → x 5.76/1.96
choice(cons(x, xs)) → choice(xs) 5.76/1.96
guess(nil) → nil 5.76/1.96
guess(cons(clause, cnf)) → cons(choice(clause), guess(cnf)) 5.76/1.96
verify(nil) → true 5.76/1.96
verify(cons(l, ls)) → if(member(negate(l), ls), false, verify(ls)) 5.76/1.96
sat(cnf) → satck(cnf, guess(cnf)) 5.76/1.96
satck(cnf, assign) → if(verify(assign), assign, unsat)

Rewrite Strategy: INNERMOST
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5.76/1.96

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
5.76/1.96
5.76/1.96

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 5.76/1.96
if(false, z0, z1) → z1 5.76/1.96
member(z0, nil) → false 5.76/1.96
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 5.76/1.96
eq(nil, nil) → true 5.76/1.96
eq(O(z0), 0(z1)) → eq(z0, z1) 5.76/1.96
eq(0(z0), 1(z1)) → false 5.76/1.96
eq(1(z0), 0(z1)) → false 5.76/1.96
eq(1(z0), 1(z1)) → eq(z0, z1) 5.76/1.96
negate(0(z0)) → 1(z0) 5.76/1.96
negate(1(z0)) → 0(z0) 5.76/1.96
choice(cons(z0, z1)) → z0 5.76/1.96
choice(cons(z0, z1)) → choice(z1) 5.76/1.96
guess(nil) → nil 5.76/1.96
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 5.76/1.96
verify(nil) → true 5.76/1.96
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 5.76/1.96
sat(z0) → satck(z0, guess(z0)) 5.76/1.96
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

MEMBER(z0, cons(z1, z2)) → c3(IF(eq(z0, z1), true, member(z0, z2)), EQ(z0, z1), MEMBER(z0, z2)) 5.76/1.96
EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 5.76/1.96
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 5.76/1.96
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 5.76/1.96
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 5.76/1.96
VERIFY(cons(z0, z1)) → c16(IF(member(negate(z0), z1), false, verify(z1)), MEMBER(negate(z0), z1), NEGATE(z0), VERIFY(z1)) 5.76/1.96
SAT(z0) → c17(SATCK(z0, guess(z0)), GUESS(z0)) 5.76/1.96
SATCK(z0, z1) → c18(IF(verify(z1), z1, unsat), VERIFY(z1))
S tuples:

MEMBER(z0, cons(z1, z2)) → c3(IF(eq(z0, z1), true, member(z0, z2)), EQ(z0, z1), MEMBER(z0, z2)) 5.76/1.99
EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 5.76/1.99
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 5.76/1.99
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 5.76/1.99
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 5.76/1.99
VERIFY(cons(z0, z1)) → c16(IF(member(negate(z0), z1), false, verify(z1)), MEMBER(negate(z0), z1), NEGATE(z0), VERIFY(z1)) 5.76/1.99
SAT(z0) → c17(SATCK(z0, guess(z0)), GUESS(z0)) 5.76/1.99
SATCK(z0, z1) → c18(IF(verify(z1), z1, unsat), VERIFY(z1))
K tuples:none
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

MEMBER, EQ, CHOICE, GUESS, VERIFY, SAT, SATCK

Compound Symbols:

c3, c5, c8, c12, c14, c16, c17, c18

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5.76/1.99

(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 4 trailing tuple parts
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5.76/1.99

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 5.76/1.99
if(false, z0, z1) → z1 5.76/1.99
member(z0, nil) → false 5.76/1.99
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 5.76/1.99
eq(nil, nil) → true 5.76/1.99
eq(O(z0), 0(z1)) → eq(z0, z1) 5.76/1.99
eq(0(z0), 1(z1)) → false 5.76/1.99
eq(1(z0), 0(z1)) → false 5.76/1.99
eq(1(z0), 1(z1)) → eq(z0, z1) 5.76/1.99
negate(0(z0)) → 1(z0) 5.76/1.99
negate(1(z0)) → 0(z0) 5.76/1.99
choice(cons(z0, z1)) → z0 5.76/1.99
choice(cons(z0, z1)) → choice(z1) 5.76/1.99
guess(nil) → nil 5.76/1.99
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 5.76/1.99
verify(nil) → true 5.76/1.99
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 5.76/1.99
sat(z0) → satck(z0, guess(z0)) 5.76/1.99
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 5.76/1.99
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 5.76/1.99
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 5.76/1.99
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 5.76/1.99
SAT(z0) → c17(SATCK(z0, guess(z0)), GUESS(z0)) 5.76/1.99
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 5.76/1.99
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 5.76/1.99
SATCK(z0, z1) → c18(VERIFY(z1))
S tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 5.76/1.99
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 5.76/1.99
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 5.76/1.99
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 5.76/1.99
SAT(z0) → c17(SATCK(z0, guess(z0)), GUESS(z0)) 5.76/1.99
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 5.76/1.99
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 5.76/1.99
SATCK(z0, z1) → c18(VERIFY(z1))
K tuples:none
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, SAT, MEMBER, VERIFY, SATCK

Compound Symbols:

c5, c8, c12, c14, c17, c3, c16, c18

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5.76/1.99

(5) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)

Split RHS of tuples not part of any SCC
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5.76/1.99

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 5.76/1.99
if(false, z0, z1) → z1 5.76/1.99
member(z0, nil) → false 5.76/1.99
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 6.15/2.02
eq(nil, nil) → true 6.15/2.02
eq(O(z0), 0(z1)) → eq(z0, z1) 6.15/2.02
eq(0(z0), 1(z1)) → false 6.15/2.02
eq(1(z0), 0(z1)) → false 6.15/2.02
eq(1(z0), 1(z1)) → eq(z0, z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
verify(nil) → true 6.15/2.02
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 6.15/2.02
sat(z0) → satck(z0, guess(z0)) 6.15/2.02
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SAT(z0) → c(GUESS(z0))
S tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SAT(z0) → c(GUESS(z0))
K tuples:none
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, MEMBER, VERIFY, SATCK, SAT

Compound Symbols:

c5, c8, c12, c14, c3, c16, c18, c

6.15/2.02
6.15/2.02

(7) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

SAT(z0) → c(GUESS(z0))
6.15/2.02
6.15/2.02

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 6.15/2.02
if(false, z0, z1) → z1 6.15/2.02
member(z0, nil) → false 6.15/2.02
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 6.15/2.02
eq(nil, nil) → true 6.15/2.02
eq(O(z0), 0(z1)) → eq(z0, z1) 6.15/2.02
eq(0(z0), 1(z1)) → false 6.15/2.02
eq(1(z0), 0(z1)) → false 6.15/2.02
eq(1(z0), 1(z1)) → eq(z0, z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
verify(nil) → true 6.15/2.02
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 6.15/2.02
sat(z0) → satck(z0, guess(z0)) 6.15/2.02
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
S tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
K tuples:none
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, MEMBER, VERIFY, SATCK, SAT

Compound Symbols:

c5, c8, c12, c14, c3, c16, c18, c

6.15/2.02
6.15/2.02

(9) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

The following tuples could be moved from S to K by knowledge propagation:

SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1))
6.15/2.02
6.15/2.02

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 6.15/2.02
if(false, z0, z1) → z1 6.15/2.02
member(z0, nil) → false 6.15/2.02
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 6.15/2.02
eq(nil, nil) → true 6.15/2.02
eq(O(z0), 0(z1)) → eq(z0, z1) 6.15/2.02
eq(0(z0), 1(z1)) → false 6.15/2.02
eq(1(z0), 0(z1)) → false 6.15/2.02
eq(1(z0), 1(z1)) → eq(z0, z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
verify(nil) → true 6.15/2.02
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 6.15/2.02
sat(z0) → satck(z0, guess(z0)) 6.15/2.02
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
S tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1))
K tuples:

SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1))
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, MEMBER, VERIFY, SATCK, SAT

Compound Symbols:

c5, c8, c12, c14, c3, c16, c18, c

6.15/2.02
6.15/2.02

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1))
We considered the (Usable) Rules:

guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0)
And the Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 6.15/2.02

POL(0(x1)) = 0    6.15/2.02
POL(1(x1)) = 0    6.15/2.02
POL(CHOICE(x1)) = [5]    6.15/2.02
POL(EQ(x1, x2)) = 0    6.15/2.02
POL(GUESS(x1)) = [4]x1    6.15/2.02
POL(MEMBER(x1, x2)) = 0    6.15/2.02
POL(O(x1)) = 0    6.15/2.02
POL(SAT(x1)) = [3] + [3]x1    6.15/2.02
POL(SATCK(x1, x2)) = [1] + x1    6.15/2.02
POL(VERIFY(x1)) = [1]    6.15/2.02
POL(c(x1)) = x1    6.15/2.02
POL(c12(x1)) = x1    6.15/2.02
POL(c14(x1, x2)) = x1 + x2    6.15/2.02
POL(c16(x1, x2)) = x1 + x2    6.15/2.02
POL(c18(x1)) = x1    6.15/2.02
POL(c3(x1, x2)) = x1 + x2    6.15/2.02
POL(c5(x1)) = x1    6.15/2.02
POL(c8(x1)) = x1    6.15/2.02
POL(choice(x1)) = [4] + [3]x1    6.15/2.02
POL(cons(x1, x2)) = [4] + x1 + x2    6.15/2.02
POL(guess(x1)) = [4]x1    6.15/2.02
POL(negate(x1)) = 0    6.15/2.02
POL(nil) = 0   
6.15/2.02
6.15/2.02

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 6.15/2.02
if(false, z0, z1) → z1 6.15/2.02
member(z0, nil) → false 6.15/2.02
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 6.15/2.02
eq(nil, nil) → true 6.15/2.02
eq(O(z0), 0(z1)) → eq(z0, z1) 6.15/2.02
eq(0(z0), 1(z1)) → false 6.15/2.02
eq(1(z0), 0(z1)) → false 6.15/2.02
eq(1(z0), 1(z1)) → eq(z0, z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
verify(nil) → true 6.15/2.02
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 6.15/2.02
sat(z0) → satck(z0, guess(z0)) 6.15/2.02
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
S tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1))
K tuples:

SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1))
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, MEMBER, VERIFY, SATCK, SAT

Compound Symbols:

c5, c8, c12, c14, c3, c16, c18, c

6.15/2.02
6.15/2.02

(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

CHOICE(cons(z0, z1)) → c12(CHOICE(z1))
We considered the (Usable) Rules:

guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0)
And the Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 6.15/2.02

POL(0(x1)) = x1    6.15/2.02
POL(1(x1)) = [2] + x1    6.15/2.02
POL(CHOICE(x1)) = [2] + [2]x1    6.15/2.02
POL(EQ(x1, x2)) = 0    6.15/2.02
POL(GUESS(x1)) = [4]x1    6.15/2.02
POL(MEMBER(x1, x2)) = 0    6.15/2.02
POL(O(x1)) = 0    6.15/2.02
POL(SAT(x1)) = [5] + [5]x1    6.15/2.02
POL(SATCK(x1, x2)) = [1] + x1 + x2    6.15/2.02
POL(VERIFY(x1)) = [1]    6.15/2.02
POL(c(x1)) = x1    6.15/2.02
POL(c12(x1)) = x1    6.15/2.02
POL(c14(x1, x2)) = x1 + x2    6.15/2.02
POL(c16(x1, x2)) = x1 + x2    6.15/2.02
POL(c18(x1)) = x1    6.15/2.02
POL(c3(x1, x2)) = x1 + x2    6.15/2.02
POL(c5(x1)) = x1    6.15/2.02
POL(c8(x1)) = x1    6.15/2.02
POL(choice(x1)) = [4] + [2]x1    6.15/2.02
POL(cons(x1, x2)) = [5] + x1 + x2    6.15/2.02
POL(guess(x1)) = [3] + [3]x1    6.15/2.02
POL(negate(x1)) = [2] + [2]x1    6.15/2.02
POL(nil) = 0   
6.15/2.02
6.15/2.02

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 6.15/2.02
if(false, z0, z1) → z1 6.15/2.02
member(z0, nil) → false 6.15/2.02
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 6.15/2.02
eq(nil, nil) → true 6.15/2.02
eq(O(z0), 0(z1)) → eq(z0, z1) 6.15/2.02
eq(0(z0), 1(z1)) → false 6.15/2.02
eq(1(z0), 0(z1)) → false 6.15/2.02
eq(1(z0), 1(z1)) → eq(z0, z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
verify(nil) → true 6.15/2.02
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 6.15/2.02
sat(z0) → satck(z0, guess(z0)) 6.15/2.02
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
S tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1))
K tuples:

SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1))
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, MEMBER, VERIFY, SATCK, SAT

Compound Symbols:

c5, c8, c12, c14, c3, c16, c18, c

6.15/2.02
6.15/2.02

(15) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1))
We considered the (Usable) Rules:

guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0)
And the Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 6.15/2.02

POL(0(x1)) = x1    6.15/2.02
POL(1(x1)) = x1    6.15/2.02
POL(CHOICE(x1)) = [2] + [2]x1    6.15/2.02
POL(EQ(x1, x2)) = 0    6.15/2.02
POL(GUESS(x1)) = [3]x1    6.15/2.02
POL(MEMBER(x1, x2)) = 0    6.15/2.02
POL(O(x1)) = 0    6.15/2.02
POL(SAT(x1)) = [5] + [5]x1    6.15/2.02
POL(SATCK(x1, x2)) = [5] + x2    6.15/2.02
POL(VERIFY(x1)) = [3] + x1    6.15/2.02
POL(c(x1)) = x1    6.15/2.02
POL(c12(x1)) = x1    6.15/2.02
POL(c14(x1, x2)) = x1 + x2    6.15/2.02
POL(c16(x1, x2)) = x1 + x2    6.15/2.02
POL(c18(x1)) = x1    6.15/2.02
POL(c3(x1, x2)) = x1 + x2    6.15/2.02
POL(c5(x1)) = x1    6.15/2.02
POL(c8(x1)) = x1    6.15/2.02
POL(choice(x1)) = [2] + [3]x1    6.15/2.02
POL(cons(x1, x2)) = [2] + x1 + x2    6.15/2.02
POL(guess(x1)) = [5]x1    6.15/2.02
POL(negate(x1)) = 0    6.15/2.02
POL(nil) = 0   
6.15/2.02
6.15/2.02

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 6.15/2.02
if(false, z0, z1) → z1 6.15/2.02
member(z0, nil) → false 6.15/2.02
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 6.15/2.02
eq(nil, nil) → true 6.15/2.02
eq(O(z0), 0(z1)) → eq(z0, z1) 6.15/2.02
eq(0(z0), 1(z1)) → false 6.15/2.02
eq(1(z0), 0(z1)) → false 6.15/2.02
eq(1(z0), 1(z1)) → eq(z0, z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
verify(nil) → true 6.15/2.02
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 6.15/2.02
sat(z0) → satck(z0, guess(z0)) 6.15/2.02
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
S tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2))
K tuples:

SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1))
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, MEMBER, VERIFY, SATCK, SAT

Compound Symbols:

c5, c8, c12, c14, c3, c16, c18, c

6.15/2.02
6.15/2.02

(17) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2))
We considered the (Usable) Rules:

guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0)
And the Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 6.15/2.02

POL(0(x1)) = 0    6.15/2.02
POL(1(x1)) = 0    6.15/2.02
POL(CHOICE(x1)) = [3]    6.15/2.02
POL(EQ(x1, x2)) = 0    6.15/2.02
POL(GUESS(x1)) = [3]x1 + [3]x12    6.15/2.02
POL(MEMBER(x1, x2)) = x2    6.15/2.02
POL(O(x1)) = 0    6.15/2.02
POL(SAT(x1)) = [3] + [3]x1 + [3]x12    6.15/2.02
POL(SATCK(x1, x2)) = [3] + x1 + x2 + [2]x22 + x1·x2    6.15/2.02
POL(VERIFY(x1)) = [3] + [2]x12    6.15/2.02
POL(c(x1)) = x1    6.15/2.02
POL(c12(x1)) = x1    6.15/2.02
POL(c14(x1, x2)) = x1 + x2    6.15/2.02
POL(c16(x1, x2)) = x1 + x2    6.15/2.02
POL(c18(x1)) = x1    6.15/2.02
POL(c3(x1, x2)) = x1 + x2    6.15/2.02
POL(c5(x1)) = x1    6.15/2.02
POL(c8(x1)) = x1    6.15/2.02
POL(choice(x1)) = [3]    6.15/2.02
POL(cons(x1, x2)) = [1] + x2    6.15/2.02
POL(guess(x1)) = x1    6.15/2.02
POL(negate(x1)) = [3]x12    6.15/2.02
POL(nil) = 0   
6.15/2.02
6.15/2.02

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 6.15/2.02
if(false, z0, z1) → z1 6.15/2.02
member(z0, nil) → false 6.15/2.02
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 6.15/2.02
eq(nil, nil) → true 6.15/2.02
eq(O(z0), 0(z1)) → eq(z0, z1) 6.15/2.02
eq(0(z0), 1(z1)) → false 6.15/2.02
eq(1(z0), 0(z1)) → false 6.15/2.02
eq(1(z0), 1(z1)) → eq(z0, z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
verify(nil) → true 6.15/2.02
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 6.15/2.02
sat(z0) → satck(z0, guess(z0)) 6.15/2.02
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
S tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1))
K tuples:

SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2))
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, MEMBER, VERIFY, SATCK, SAT

Compound Symbols:

c5, c8, c12, c14, c3, c16, c18, c

6.15/2.02
6.15/2.02

(19) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

EQ(1(z0), 1(z1)) → c8(EQ(z0, z1))
We considered the (Usable) Rules:

guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0)
And the Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 6.15/2.02

POL(0(x1)) = x1    6.15/2.02
POL(1(x1)) = [1] + x1    6.15/2.02
POL(CHOICE(x1)) = [3] + x1 + [3]x12    6.15/2.02
POL(EQ(x1, x2)) = [2] + [2]x2    6.15/2.02
POL(GUESS(x1)) = [3]x1 + [3]x12    6.15/2.02
POL(MEMBER(x1, x2)) = [2]x2    6.15/2.02
POL(O(x1)) = 0    6.15/2.02
POL(SAT(x1)) = [3] + [3]x1 + [2]x12    6.15/2.02
POL(SATCK(x1, x2)) = [2] + x1 + x22 + x12    6.15/2.02
POL(VERIFY(x1)) = [2] + x12    6.15/2.02
POL(c(x1)) = x1    6.15/2.02
POL(c12(x1)) = x1    6.15/2.02
POL(c14(x1, x2)) = x1 + x2    6.15/2.02
POL(c16(x1, x2)) = x1 + x2    6.15/2.02
POL(c18(x1)) = x1    6.15/2.02
POL(c3(x1, x2)) = x1 + x2    6.15/2.02
POL(c5(x1)) = x1    6.15/2.02
POL(c8(x1)) = x1    6.15/2.02
POL(choice(x1)) = x1    6.15/2.02
POL(cons(x1, x2)) = [1] + x1 + x2    6.15/2.02
POL(guess(x1)) = [1] + x1    6.15/2.02
POL(negate(x1)) = [3]x12    6.15/2.02
POL(nil) = 0   
6.15/2.02
6.15/2.02

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 6.15/2.02
if(false, z0, z1) → z1 6.15/2.02
member(z0, nil) → false 6.15/2.02
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 6.15/2.02
eq(nil, nil) → true 6.15/2.02
eq(O(z0), 0(z1)) → eq(z0, z1) 6.15/2.02
eq(0(z0), 1(z1)) → false 6.15/2.02
eq(1(z0), 0(z1)) → false 6.15/2.02
eq(1(z0), 1(z1)) → eq(z0, z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
verify(nil) → true 6.15/2.02
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 6.15/2.02
sat(z0) → satck(z0, guess(z0)) 6.15/2.02
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
S tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1))
K tuples:

SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1))
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, MEMBER, VERIFY, SATCK, SAT

Compound Symbols:

c5, c8, c12, c14, c3, c16, c18, c

6.15/2.02
6.15/2.02

(21) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1))
We considered the (Usable) Rules:

guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0)
And the Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 6.15/2.02

POL(0(x1)) = [1] + x1    6.15/2.02
POL(1(x1)) = x1    6.15/2.02
POL(CHOICE(x1)) = [2] + [3]x1 + [3]x12    6.15/2.02
POL(EQ(x1, x2)) = [2] + [2]x2    6.15/2.02
POL(GUESS(x1)) = [3]x1 + [3]x12    6.15/2.02
POL(MEMBER(x1, x2)) = [3] + [2]x2    6.15/2.02
POL(O(x1)) = 0    6.15/2.02
POL(SAT(x1)) = [3] + [3]x1 + [3]x12    6.15/2.02
POL(SATCK(x1, x2)) = [3] + x1 + [2]x2 + [2]x22    6.15/2.02
POL(VERIFY(x1)) = [3] + [2]x1 + x12    6.15/2.02
POL(c(x1)) = x1    6.15/2.02
POL(c12(x1)) = x1    6.15/2.02
POL(c14(x1, x2)) = x1 + x2    6.15/2.02
POL(c16(x1, x2)) = x1 + x2    6.15/2.02
POL(c18(x1)) = x1    6.15/2.02
POL(c3(x1, x2)) = x1 + x2    6.15/2.02
POL(c5(x1)) = x1    6.15/2.02
POL(c8(x1)) = x1    6.15/2.02
POL(choice(x1)) = x1    6.15/2.02
POL(cons(x1, x2)) = [1] + x1 + x2    6.15/2.02
POL(guess(x1)) = x1    6.15/2.02
POL(negate(x1)) = [3]x12    6.15/2.02
POL(nil) = 0   
6.15/2.02
6.15/2.02

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

if(true, z0, z1) → z0 6.15/2.02
if(false, z0, z1) → z1 6.15/2.02
member(z0, nil) → false 6.15/2.02
member(z0, cons(z1, z2)) → if(eq(z0, z1), true, member(z0, z2)) 6.15/2.02
eq(nil, nil) → true 6.15/2.02
eq(O(z0), 0(z1)) → eq(z0, z1) 6.15/2.02
eq(0(z0), 1(z1)) → false 6.15/2.02
eq(1(z0), 0(z1)) → false 6.15/2.02
eq(1(z0), 1(z1)) → eq(z0, z1) 6.15/2.02
negate(0(z0)) → 1(z0) 6.15/2.02
negate(1(z0)) → 0(z0) 6.15/2.02
choice(cons(z0, z1)) → z0 6.15/2.02
choice(cons(z0, z1)) → choice(z1) 6.15/2.02
guess(nil) → nil 6.15/2.02
guess(cons(z0, z1)) → cons(choice(z0), guess(z1)) 6.15/2.02
verify(nil) → true 6.15/2.02
verify(cons(z0, z1)) → if(member(negate(z0), z1), false, verify(z1)) 6.15/2.02
sat(z0) → satck(z0, guess(z0)) 6.15/2.02
satck(z0, z1) → if(verify(z1), z1, unsat)
Tuples:

EQ(O(z0), 0(z1)) → c5(EQ(z0, z1)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
SAT(z0) → c(SATCK(z0, guess(z0)))
S tuples:none
K tuples:

SAT(z0) → c(SATCK(z0, guess(z0))) 6.15/2.02
SATCK(z0, z1) → c18(VERIFY(z1)) 6.15/2.02
GUESS(cons(z0, z1)) → c14(CHOICE(z0), GUESS(z1)) 6.15/2.02
CHOICE(cons(z0, z1)) → c12(CHOICE(z1)) 6.15/2.02
VERIFY(cons(z0, z1)) → c16(MEMBER(negate(z0), z1), VERIFY(z1)) 6.15/2.02
MEMBER(z0, cons(z1, z2)) → c3(EQ(z0, z1), MEMBER(z0, z2)) 6.15/2.02
EQ(1(z0), 1(z1)) → c8(EQ(z0, z1)) 6.15/2.02
EQ(O(z0), 0(z1)) → c5(EQ(z0, z1))
Defined Rule Symbols:

if, member, eq, negate, choice, guess, verify, sat, satck

Defined Pair Symbols:

EQ, CHOICE, GUESS, MEMBER, VERIFY, SATCK, SAT

Compound Symbols:

c5, c8, c12, c14, c3, c16, c18, c

6.15/2.02
6.15/2.02

(23) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty
6.15/2.02
6.15/2.02

(24) BOUNDS(O(1), O(1))

6.15/2.02
6.15/2.02
6.15/2.07 EOF