YES(O(1), O(n^1)) 2.34/1.03 YES(O(1), O(n^1)) 2.34/1.07 2.34/1.07 2.34/1.07 2.34/1.07 2.34/1.07 2.34/1.07 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 2.34/1.07 2.34/1.07 2.34/1.07
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
2.34/1.07 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
2.34/1.07 
↳2 CdtProblem
2.34/1.07 
↳3 CdtUnreachableProof (⇔)
2.34/1.07 
↳4 CdtProblem
2.34/1.07 
↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
2.34/1.07 
↳6 CdtProblem
2.34/1.07 
↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
2.34/1.07 
↳8 CdtProblem
2.34/1.07 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
2.34/1.07 
↳10 CdtProblem
2.34/1.07 
↳11 CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
2.34/1.07 
↳12 CdtProblem
2.34/1.07 
↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
2.34/1.07 
↳14 CdtProblem
2.34/1.07 
↳15 SIsEmptyProof (BOTH BOUNDS(ID, ID))
2.34/1.07 
↳16 BOUNDS(O(1), O(1))
2.34/1.07
2.34/1.07 2.34/1.07
2.34/1.07
2.34/1.07

(0) Obligation:

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

from(X) → cons(X, n__from(n__s(X))) 2.34/1.07
2ndspos(0, Z) → rnil 2.34/1.07
2ndspos(s(N), cons(X, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z))) 2.34/1.07
2ndsneg(0, Z) → rnil 2.34/1.07
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z))) 2.34/1.07
pi(X) → 2ndspos(X, from(0)) 2.34/1.07
plus(0, Y) → Y 2.34/1.07
plus(s(X), Y) → s(plus(X, Y)) 2.34/1.07
times(0, Y) → 0 2.34/1.07
times(s(X), Y) → plus(Y, times(X, Y)) 2.34/1.07
square(X) → times(X, X) 2.34/1.07
from(X) → n__from(X) 2.34/1.07
s(X) → n__s(X) 2.34/1.07
cons(X1, X2) → n__cons(X1, X2) 2.34/1.07
activate(n__from(X)) → from(activate(X)) 2.34/1.07
activate(n__s(X)) → s(activate(X)) 2.34/1.07
activate(n__cons(X1, X2)) → cons(activate(X1), X2) 2.34/1.07
activate(X) → X

Rewrite Strategy: INNERMOST
2.34/1.07
2.34/1.07

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

Converted CpxTRS to CDT
2.34/1.07
2.34/1.07

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0))) 2.34/1.07
from(z0) → n__from(z0) 2.34/1.07
2ndspos(0, z0) → rnil 2.34/1.07
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3))) 2.34/1.07
2ndsneg(0, z0) → rnil 2.34/1.07
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3))) 2.34/1.07
pi(z0) → 2ndspos(z0, from(0)) 2.34/1.07
plus(0, z0) → z0 2.34/1.07
plus(s(z0), z1) → s(plus(z0, z1)) 2.34/1.07
times(0, z0) → 0 2.34/1.07
times(s(z0), z1) → plus(z1, times(z0, z1)) 2.34/1.07
square(z0) → times(z0, z0) 2.34/1.07
s(z0) → n__s(z0) 2.34/1.07
cons(z0, z1) → n__cons(z0, z1) 2.34/1.07
activate(n__from(z0)) → from(activate(z0)) 2.34/1.07
activate(n__s(z0)) → s(activate(z0)) 2.34/1.07
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 2.34/1.07
activate(z0) → z0
Tuples:

FROM(z0) → c(CONS(z0, n__from(n__s(z0)))) 2.34/1.07
2NDSPOS(s(z0), cons(z1, n__cons(z2, z3))) → c3(ACTIVATE(z2), 2NDSNEG(z0, activate(z3)), ACTIVATE(z3)) 2.34/1.07
2NDSNEG(s(z0), cons(z1, n__cons(z2, z3))) → c5(ACTIVATE(z2), 2NDSPOS(z0, activate(z3)), ACTIVATE(z3)) 2.34/1.07
PI(z0) → c6(2NDSPOS(z0, from(0)), FROM(0)) 2.34/1.07
PLUS(s(z0), z1) → c8(S(plus(z0, z1)), PLUS(z0, z1)) 2.34/1.07
TIMES(s(z0), z1) → c10(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 2.34/1.07
SQUARE(z0) → c11(TIMES(z0, z0)) 2.34/1.07
ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))
S tuples:

FROM(z0) → c(CONS(z0, n__from(n__s(z0)))) 2.34/1.07
2NDSPOS(s(z0), cons(z1, n__cons(z2, z3))) → c3(ACTIVATE(z2), 2NDSNEG(z0, activate(z3)), ACTIVATE(z3)) 2.34/1.07
2NDSNEG(s(z0), cons(z1, n__cons(z2, z3))) → c5(ACTIVATE(z2), 2NDSPOS(z0, activate(z3)), ACTIVATE(z3)) 2.34/1.07
PI(z0) → c6(2NDSPOS(z0, from(0)), FROM(0)) 2.34/1.07
PLUS(s(z0), z1) → c8(S(plus(z0, z1)), PLUS(z0, z1)) 2.34/1.07
TIMES(s(z0), z1) → c10(PLUS(z1, times(z0, z1)), TIMES(z0, z1)) 2.34/1.07
SQUARE(z0) → c11(TIMES(z0, z0)) 2.34/1.07
ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

FROM, 2NDSPOS, 2NDSNEG, PI, PLUS, TIMES, SQUARE, ACTIVATE

Compound Symbols:

c, c3, c5, c6, c8, c10, c11, c14, c15, c16

2.34/1.07
2.34/1.07

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

2NDSPOS(s(z0), cons(z1, n__cons(z2, z3))) → c3(ACTIVATE(z2), 2NDSNEG(z0, activate(z3)), ACTIVATE(z3)) 2.34/1.07
2NDSNEG(s(z0), cons(z1, n__cons(z2, z3))) → c5(ACTIVATE(z2), 2NDSPOS(z0, activate(z3)), ACTIVATE(z3)) 2.34/1.07
PLUS(s(z0), z1) → c8(S(plus(z0, z1)), PLUS(z0, z1)) 2.34/1.07
TIMES(s(z0), z1) → c10(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
2.34/1.07
2.34/1.07

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0))) 2.34/1.07
from(z0) → n__from(z0) 2.34/1.07
2ndspos(0, z0) → rnil 2.34/1.07
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3))) 2.34/1.07
2ndsneg(0, z0) → rnil 2.34/1.07
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3))) 2.34/1.07
pi(z0) → 2ndspos(z0, from(0)) 2.34/1.07
plus(0, z0) → z0 2.34/1.07
plus(s(z0), z1) → s(plus(z0, z1)) 2.34/1.07
times(0, z0) → 0 2.34/1.07
times(s(z0), z1) → plus(z1, times(z0, z1)) 2.34/1.07
square(z0) → times(z0, z0) 2.34/1.07
s(z0) → n__s(z0) 2.34/1.07
cons(z0, z1) → n__cons(z0, z1) 2.34/1.07
activate(n__from(z0)) → from(activate(z0)) 2.34/1.07
activate(n__s(z0)) → s(activate(z0)) 2.34/1.07
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 2.34/1.07
activate(z0) → z0
Tuples:

FROM(z0) → c(CONS(z0, n__from(n__s(z0)))) 2.34/1.07
PI(z0) → c6(2NDSPOS(z0, from(0)), FROM(0)) 2.34/1.07
SQUARE(z0) → c11(TIMES(z0, z0)) 2.34/1.07
ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))
S tuples:

FROM(z0) → c(CONS(z0, n__from(n__s(z0)))) 2.34/1.07
PI(z0) → c6(2NDSPOS(z0, from(0)), FROM(0)) 2.34/1.07
SQUARE(z0) → c11(TIMES(z0, z0)) 2.34/1.07
ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

FROM, PI, SQUARE, ACTIVATE

Compound Symbols:

c, c6, c11, c14, c15, c16

2.34/1.07
2.34/1.07

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

Removed 5 trailing tuple parts
2.34/1.07
2.34/1.07

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0))) 2.34/1.07
from(z0) → n__from(z0) 2.34/1.07
2ndspos(0, z0) → rnil 2.34/1.07
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3))) 2.34/1.07
2ndsneg(0, z0) → rnil 2.34/1.07
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3))) 2.34/1.07
pi(z0) → 2ndspos(z0, from(0)) 2.34/1.07
plus(0, z0) → z0 2.34/1.07
plus(s(z0), z1) → s(plus(z0, z1)) 2.34/1.07
times(0, z0) → 0 2.34/1.07
times(s(z0), z1) → plus(z1, times(z0, z1)) 2.34/1.07
square(z0) → times(z0, z0) 2.34/1.07
s(z0) → n__s(z0) 2.34/1.07
cons(z0, z1) → n__cons(z0, z1) 2.34/1.07
activate(n__from(z0)) → from(activate(z0)) 2.34/1.07
activate(n__s(z0)) → s(activate(z0)) 2.34/1.07
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 2.34/1.07
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.07
FROM(z0) → c 2.34/1.07
PI(z0) → c6(FROM(0)) 2.34/1.07
SQUARE(z0) → c11 2.34/1.07
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
S tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.07
FROM(z0) → c 2.34/1.07
PI(z0) → c6(FROM(0)) 2.34/1.07
SQUARE(z0) → c11 2.34/1.07
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

ACTIVATE, FROM, PI, SQUARE

Compound Symbols:

c14, c, c6, c11, c15, c16

2.34/1.07
2.34/1.07

(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing nodes:

FROM(z0) → c 2.34/1.07
PI(z0) → c6(FROM(0)) 2.34/1.07
SQUARE(z0) → c11
2.34/1.07
2.34/1.07

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0))) 2.34/1.07
from(z0) → n__from(z0) 2.34/1.07
2ndspos(0, z0) → rnil 2.34/1.07
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3))) 2.34/1.07
2ndsneg(0, z0) → rnil 2.34/1.07
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3))) 2.34/1.07
pi(z0) → 2ndspos(z0, from(0)) 2.34/1.07
plus(0, z0) → z0 2.34/1.07
plus(s(z0), z1) → s(plus(z0, z1)) 2.34/1.07
times(0, z0) → 0 2.34/1.07
times(s(z0), z1) → plus(z1, times(z0, z1)) 2.34/1.07
square(z0) → times(z0, z0) 2.34/1.07
s(z0) → n__s(z0) 2.34/1.07
cons(z0, z1) → n__cons(z0, z1) 2.34/1.07
activate(n__from(z0)) → from(activate(z0)) 2.34/1.07
activate(n__s(z0)) → s(activate(z0)) 2.34/1.07
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 2.34/1.07
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.07
FROM(z0) → c 2.34/1.07
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
S tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.07
FROM(z0) → c 2.34/1.07
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0)) 2.34/1.07
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

ACTIVATE, FROM

Compound Symbols:

c14, c, c15, c16

2.34/1.07
2.34/1.07

(9) 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.

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.08
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
We considered the (Usable) Rules:

activate(n__from(z0)) → from(activate(z0)) 2.34/1.08
activate(n__s(z0)) → s(activate(z0)) 2.34/1.08
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 2.34/1.08
activate(z0) → z0 2.34/1.08
cons(z0, z1) → n__cons(z0, z1) 2.34/1.08
s(z0) → n__s(z0) 2.34/1.08
from(z0) → cons(z0, n__from(n__s(z0))) 2.34/1.08
from(z0) → n__from(z0)
And the Tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.08
FROM(z0) → c 2.34/1.08
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0)) 2.34/1.08
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.34/1.08

POL(ACTIVATE(x1)) = [2]x1    2.34/1.08
POL(FROM(x1)) = 0    2.34/1.08
POL(activate(x1)) = 0    2.34/1.08
POL(c) = 0    2.34/1.08
POL(c14(x1, x2)) = x1 + x2    2.34/1.08
POL(c15(x1)) = x1    2.34/1.08
POL(c16(x1)) = x1    2.34/1.08
POL(cons(x1, x2)) = [3] + [3]x1    2.34/1.08
POL(from(x1)) = [3] + [3]x1    2.34/1.08
POL(n__cons(x1, x2)) = [1] + x1 + x2    2.34/1.08
POL(n__from(x1)) = [1] + x1    2.34/1.08
POL(n__s(x1)) = x1    2.34/1.08
POL(s(x1)) = [3] + [3]x1   
2.34/1.08
2.34/1.08

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0))) 2.34/1.08
from(z0) → n__from(z0) 2.34/1.08
2ndspos(0, z0) → rnil 2.34/1.08
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3))) 2.34/1.08
2ndsneg(0, z0) → rnil 2.34/1.08
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3))) 2.34/1.08
pi(z0) → 2ndspos(z0, from(0)) 2.34/1.08
plus(0, z0) → z0 2.34/1.08
plus(s(z0), z1) → s(plus(z0, z1)) 2.34/1.08
times(0, z0) → 0 2.34/1.08
times(s(z0), z1) → plus(z1, times(z0, z1)) 2.34/1.08
square(z0) → times(z0, z0) 2.34/1.08
s(z0) → n__s(z0) 2.34/1.08
cons(z0, z1) → n__cons(z0, z1) 2.34/1.08
activate(n__from(z0)) → from(activate(z0)) 2.34/1.08
activate(n__s(z0)) → s(activate(z0)) 2.34/1.08
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 2.34/1.08
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.08
FROM(z0) → c 2.34/1.08
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0)) 2.34/1.08
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
S tuples:

FROM(z0) → c 2.34/1.08
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
K tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.08
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

ACTIVATE, FROM

Compound Symbols:

c14, c, c15, c16

2.34/1.08
2.34/1.08

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

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

FROM(z0) → c
2.34/1.08
2.34/1.08

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0))) 2.34/1.08
from(z0) → n__from(z0) 2.34/1.08
2ndspos(0, z0) → rnil 2.34/1.08
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3))) 2.34/1.08
2ndsneg(0, z0) → rnil 2.34/1.08
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3))) 2.34/1.08
pi(z0) → 2ndspos(z0, from(0)) 2.34/1.08
plus(0, z0) → z0 2.34/1.08
plus(s(z0), z1) → s(plus(z0, z1)) 2.34/1.08
times(0, z0) → 0 2.34/1.08
times(s(z0), z1) → plus(z1, times(z0, z1)) 2.34/1.08
square(z0) → times(z0, z0) 2.34/1.08
s(z0) → n__s(z0) 2.34/1.08
cons(z0, z1) → n__cons(z0, z1) 2.34/1.08
activate(n__from(z0)) → from(activate(z0)) 2.34/1.08
activate(n__s(z0)) → s(activate(z0)) 2.34/1.08
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 2.34/1.08
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.08
FROM(z0) → c 2.34/1.08
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0)) 2.34/1.08
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
S tuples:

ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
K tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.08
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0)) 2.34/1.08
FROM(z0) → c
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

ACTIVATE, FROM

Compound Symbols:

c14, c, c15, c16

2.34/1.08
2.34/1.08

(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.

ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
We considered the (Usable) Rules:

activate(n__from(z0)) → from(activate(z0)) 2.34/1.08
activate(n__s(z0)) → s(activate(z0)) 2.34/1.08
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 2.34/1.08
activate(z0) → z0 2.34/1.08
cons(z0, z1) → n__cons(z0, z1) 2.34/1.08
s(z0) → n__s(z0) 2.34/1.08
from(z0) → cons(z0, n__from(n__s(z0))) 2.34/1.08
from(z0) → n__from(z0)
And the Tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.08
FROM(z0) → c 2.34/1.08
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0)) 2.34/1.08
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 2.34/1.08

POL(ACTIVATE(x1)) = [4]x1    2.34/1.08
POL(FROM(x1)) = [3]    2.34/1.08
POL(activate(x1)) = [2]x1    2.34/1.08
POL(c) = 0    2.34/1.08
POL(c14(x1, x2)) = x1 + x2    2.34/1.08
POL(c15(x1)) = x1    2.34/1.08
POL(c16(x1)) = x1    2.34/1.08
POL(cons(x1, x2)) = x1    2.34/1.08
POL(from(x1)) = [2] + x1    2.34/1.08
POL(n__cons(x1, x2)) = x1    2.34/1.08
POL(n__from(x1)) = [2] + x1    2.34/1.08
POL(n__s(x1)) = [4] + x1    2.34/1.08
POL(s(x1)) = [4] + x1   
2.34/1.08
2.34/1.08

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0))) 2.34/1.08
from(z0) → n__from(z0) 2.34/1.08
2ndspos(0, z0) → rnil 2.34/1.08
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3))) 2.34/1.08
2ndsneg(0, z0) → rnil 2.34/1.08
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3))) 2.34/1.08
pi(z0) → 2ndspos(z0, from(0)) 2.34/1.08
plus(0, z0) → z0 2.34/1.08
plus(s(z0), z1) → s(plus(z0, z1)) 2.34/1.08
times(0, z0) → 0 2.34/1.08
times(s(z0), z1) → plus(z1, times(z0, z1)) 2.34/1.08
square(z0) → times(z0, z0) 2.34/1.08
s(z0) → n__s(z0) 2.34/1.08
cons(z0, z1) → n__cons(z0, z1) 2.34/1.08
activate(n__from(z0)) → from(activate(z0)) 2.34/1.08
activate(n__s(z0)) → s(activate(z0)) 2.34/1.08
activate(n__cons(z0, z1)) → cons(activate(z0), z1) 2.34/1.08
activate(z0) → z0
Tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.08
FROM(z0) → c 2.34/1.08
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0)) 2.34/1.08
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))
S tuples:none
K tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0)) 2.34/1.08
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0)) 2.34/1.08
FROM(z0) → c 2.34/1.08
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

ACTIVATE, FROM

Compound Symbols:

c14, c, c15, c16

2.34/1.08
2.34/1.08

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

The set S is empty
2.34/1.08
2.34/1.08

(16) BOUNDS(O(1), O(1))

2.34/1.08
2.34/1.08
2.78/1.26 EOF