YES(O(1), O(n^1)) 0.00/0.93 YES(O(1), O(n^1)) 0.00/0.97 0.00/0.97 0.00/0.97 0.00/0.97 0.00/0.97 0.00/0.97 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.97 0.00/0.97 0.00/0.97
0.00/0.97
0.00/0.97
0.00/0.97
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
0.00/0.97 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
0.00/0.97 
↳2 CdtProblem
0.00/0.97 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
0.00/0.97 
↳4 CdtProblem
0.00/0.97 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.97 
↳6 CdtProblem
0.00/0.97 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.97 
↳8 CdtProblem
0.00/0.97 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.97 
↳10 CdtProblem
0.00/0.97 
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/0.97 
↳12 BOUNDS(O(1), O(1))
0.00/0.97
0.00/0.97 0.00/0.97
0.00/0.97
0.00/0.97

(0) Obligation:

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

a__eq(0, 0) → true 0.00/0.97
a__eq(s(X), s(Y)) → a__eq(X, Y) 0.00/0.97
a__eq(X, Y) → false 0.00/0.97
a__inf(X) → cons(X, inf(s(X))) 0.00/0.97
a__take(0, X) → nil 0.00/0.97
a__take(s(X), cons(Y, L)) → cons(Y, take(X, L)) 0.00/0.97
a__length(nil) → 0 0.00/0.97
a__length(cons(X, L)) → s(length(L)) 0.00/0.97
mark(eq(X1, X2)) → a__eq(X1, X2) 0.00/0.97
mark(inf(X)) → a__inf(mark(X)) 0.00/0.97
mark(take(X1, X2)) → a__take(mark(X1), mark(X2)) 0.00/0.97
mark(length(X)) → a__length(mark(X)) 0.00/0.97
mark(0) → 0 0.00/0.97
mark(true) → true 0.00/0.97
mark(s(X)) → s(X) 0.00/0.97
mark(false) → false 0.00/0.97
mark(cons(X1, X2)) → cons(X1, X2) 0.00/0.97
mark(nil) → nil 0.00/0.97
a__eq(X1, X2) → eq(X1, X2) 0.00/0.97
a__inf(X) → inf(X) 0.00/0.97
a__take(X1, X2) → take(X1, X2) 0.00/0.97
a__length(X) → length(X)

Rewrite Strategy: INNERMOST
0.00/0.97
0.00/0.97

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

Converted CpxTRS to CDT
0.00/0.97
0.00/0.97

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true 0.00/0.97
a__eq(s(z0), s(z1)) → a__eq(z0, z1) 0.00/0.97
a__eq(z0, z1) → false 0.00/0.97
a__eq(z0, z1) → eq(z0, z1) 0.00/0.97
a__inf(z0) → cons(z0, inf(s(z0))) 0.00/0.97
a__inf(z0) → inf(z0) 0.00/0.97
a__take(0, z0) → nil 0.00/0.97
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2)) 0.00/0.97
a__take(z0, z1) → take(z0, z1) 0.00/0.97
a__length(nil) → 0 0.00/0.97
a__length(cons(z0, z1)) → s(length(z1)) 0.00/0.97
a__length(z0) → length(z0) 0.00/0.97
mark(eq(z0, z1)) → a__eq(z0, z1) 0.00/0.97
mark(inf(z0)) → a__inf(mark(z0)) 0.00/0.97
mark(take(z0, z1)) → a__take(mark(z0), mark(z1)) 0.00/0.97
mark(length(z0)) → a__length(mark(z0)) 0.00/0.97
mark(0) → 0 0.00/0.97
mark(true) → true 0.00/0.97
mark(s(z0)) → s(z0) 0.00/0.97
mark(false) → false 0.00/0.97
mark(cons(z0, z1)) → cons(z0, z1) 0.00/0.97
mark(nil) → nil
Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.97
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.97
MARK(inf(z0)) → c13(A__INF(mark(z0)), MARK(z0)) 0.00/0.97
MARK(take(z0, z1)) → c14(A__TAKE(mark(z0), mark(z1)), MARK(z0), MARK(z1)) 0.00/0.97
MARK(length(z0)) → c15(A__LENGTH(mark(z0)), MARK(z0))
S tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.97
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.97
MARK(inf(z0)) → c13(A__INF(mark(z0)), MARK(z0)) 0.00/0.97
MARK(take(z0, z1)) → c14(A__TAKE(mark(z0), mark(z1)), MARK(z0), MARK(z1)) 0.00/0.97
MARK(length(z0)) → c15(A__LENGTH(mark(z0)), MARK(z0))
K tuples:none
Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

0.00/0.97
0.00/0.97

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

Removed 3 trailing tuple parts
0.00/0.97
0.00/0.97

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true 0.00/0.97
a__eq(s(z0), s(z1)) → a__eq(z0, z1) 0.00/0.97
a__eq(z0, z1) → false 0.00/0.97
a__eq(z0, z1) → eq(z0, z1) 0.00/0.97
a__inf(z0) → cons(z0, inf(s(z0))) 0.00/0.97
a__inf(z0) → inf(z0) 0.00/0.97
a__take(0, z0) → nil 0.00/0.97
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2)) 0.00/0.97
a__take(z0, z1) → take(z0, z1) 0.00/0.97
a__length(nil) → 0 0.00/0.97
a__length(cons(z0, z1)) → s(length(z1)) 0.00/0.97
a__length(z0) → length(z0) 0.00/0.97
mark(eq(z0, z1)) → a__eq(z0, z1) 0.00/0.97
mark(inf(z0)) → a__inf(mark(z0)) 0.00/0.97
mark(take(z0, z1)) → a__take(mark(z0), mark(z1)) 0.00/0.97
mark(length(z0)) → a__length(mark(z0)) 0.00/0.97
mark(0) → 0 0.00/0.97
mark(true) → true 0.00/0.97
mark(s(z0)) → s(z0) 0.00/0.97
mark(false) → false 0.00/0.97
mark(cons(z0, z1)) → cons(z0, z1) 0.00/0.97
mark(nil) → nil
Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.97
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.97
MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.97
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.97
MARK(length(z0)) → c15(MARK(z0))
S tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.97
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.97
MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.97
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.97
MARK(length(z0)) → c15(MARK(z0))
K tuples:none
Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

0.00/0.98
0.00/0.98

(5) 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.

MARK(inf(z0)) → c13(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.98

POL(A__EQ(x1, x2)) = 0    0.00/0.98
POL(MARK(x1)) = x1    0.00/0.98
POL(c1(x1)) = x1    0.00/0.98
POL(c12(x1)) = x1    0.00/0.98
POL(c13(x1)) = x1    0.00/0.98
POL(c14(x1, x2)) = x1 + x2    0.00/0.98
POL(c15(x1)) = x1    0.00/0.98
POL(eq(x1, x2)) = 0    0.00/0.98
POL(inf(x1)) = [1] + x1    0.00/0.98
POL(length(x1)) = x1    0.00/0.98
POL(s(x1)) = 0    0.00/0.98
POL(take(x1, x2)) = x1 + x2   
0.00/0.98
0.00/0.98

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true 0.00/0.98
a__eq(s(z0), s(z1)) → a__eq(z0, z1) 0.00/0.98
a__eq(z0, z1) → false 0.00/0.98
a__eq(z0, z1) → eq(z0, z1) 0.00/0.98
a__inf(z0) → cons(z0, inf(s(z0))) 0.00/0.98
a__inf(z0) → inf(z0) 0.00/0.98
a__take(0, z0) → nil 0.00/0.98
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2)) 0.00/0.98
a__take(z0, z1) → take(z0, z1) 0.00/0.98
a__length(nil) → 0 0.00/0.98
a__length(cons(z0, z1)) → s(length(z1)) 0.00/0.98
a__length(z0) → length(z0) 0.00/0.98
mark(eq(z0, z1)) → a__eq(z0, z1) 0.00/0.98
mark(inf(z0)) → a__inf(mark(z0)) 0.00/0.98
mark(take(z0, z1)) → a__take(mark(z0), mark(z1)) 0.00/0.98
mark(length(z0)) → a__length(mark(z0)) 0.00/0.98
mark(0) → 0 0.00/0.98
mark(true) → true 0.00/0.98
mark(s(z0)) → s(z0) 0.00/0.98
mark(false) → false 0.00/0.98
mark(cons(z0, z1)) → cons(z0, z1) 0.00/0.98
mark(nil) → nil
Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
S tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
K tuples:

MARK(inf(z0)) → c13(MARK(z0))
Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

0.00/0.98
0.00/0.98

(7) 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.

MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
We considered the (Usable) Rules:none
And the Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.98

POL(A__EQ(x1, x2)) = [1]    0.00/0.98
POL(MARK(x1)) = x1    0.00/0.98
POL(c1(x1)) = x1    0.00/0.98
POL(c12(x1)) = x1    0.00/0.98
POL(c13(x1)) = x1    0.00/0.98
POL(c14(x1, x2)) = x1 + x2    0.00/0.98
POL(c15(x1)) = x1    0.00/0.98
POL(eq(x1, x2)) = [2]    0.00/0.98
POL(inf(x1)) = x1    0.00/0.98
POL(length(x1)) = x1    0.00/0.98
POL(s(x1)) = x1    0.00/0.98
POL(take(x1, x2)) = [2] + x1 + x2   
0.00/0.98
0.00/0.98

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true 0.00/0.98
a__eq(s(z0), s(z1)) → a__eq(z0, z1) 0.00/0.98
a__eq(z0, z1) → false 0.00/0.98
a__eq(z0, z1) → eq(z0, z1) 0.00/0.98
a__inf(z0) → cons(z0, inf(s(z0))) 0.00/0.98
a__inf(z0) → inf(z0) 0.00/0.98
a__take(0, z0) → nil 0.00/0.98
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2)) 0.00/0.98
a__take(z0, z1) → take(z0, z1) 0.00/0.98
a__length(nil) → 0 0.00/0.98
a__length(cons(z0, z1)) → s(length(z1)) 0.00/0.98
a__length(z0) → length(z0) 0.00/0.98
mark(eq(z0, z1)) → a__eq(z0, z1) 0.00/0.98
mark(inf(z0)) → a__inf(mark(z0)) 0.00/0.98
mark(take(z0, z1)) → a__take(mark(z0), mark(z1)) 0.00/0.98
mark(length(z0)) → a__length(mark(z0)) 0.00/0.98
mark(0) → 0 0.00/0.98
mark(true) → true 0.00/0.98
mark(s(z0)) → s(z0) 0.00/0.98
mark(false) → false 0.00/0.98
mark(cons(z0, z1)) → cons(z0, z1) 0.00/0.98
mark(nil) → nil
Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
S tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
K tuples:

MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.98
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1))
Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

0.00/0.98
0.00/0.98

(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.

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
We considered the (Usable) Rules:none
And the Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.98

POL(A__EQ(x1, x2)) = [5] + x2    0.00/0.98
POL(MARK(x1)) = [2] + x1    0.00/0.98
POL(c1(x1)) = x1    0.00/0.98
POL(c12(x1)) = x1    0.00/0.98
POL(c13(x1)) = x1    0.00/0.98
POL(c14(x1, x2)) = x1 + x2    0.00/0.98
POL(c15(x1)) = x1    0.00/0.98
POL(eq(x1, x2)) = [3] + x2    0.00/0.98
POL(inf(x1)) = x1    0.00/0.98
POL(length(x1)) = [1] + x1    0.00/0.98
POL(s(x1)) = [2] + x1    0.00/0.98
POL(take(x1, x2)) = [3] + x1 + x2   
0.00/0.98
0.00/0.98

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

a__eq(0, 0) → true 0.00/0.98
a__eq(s(z0), s(z1)) → a__eq(z0, z1) 0.00/0.98
a__eq(z0, z1) → false 0.00/0.98
a__eq(z0, z1) → eq(z0, z1) 0.00/0.98
a__inf(z0) → cons(z0, inf(s(z0))) 0.00/0.98
a__inf(z0) → inf(z0) 0.00/0.98
a__take(0, z0) → nil 0.00/0.98
a__take(s(z0), cons(z1, z2)) → cons(z1, take(z0, z2)) 0.00/0.98
a__take(z0, z1) → take(z0, z1) 0.00/0.98
a__length(nil) → 0 0.00/0.98
a__length(cons(z0, z1)) → s(length(z1)) 0.00/0.98
a__length(z0) → length(z0) 0.00/0.98
mark(eq(z0, z1)) → a__eq(z0, z1) 0.00/0.98
mark(inf(z0)) → a__inf(mark(z0)) 0.00/0.98
mark(take(z0, z1)) → a__take(mark(z0), mark(z1)) 0.00/0.98
mark(length(z0)) → a__length(mark(z0)) 0.00/0.98
mark(0) → 0 0.00/0.98
mark(true) → true 0.00/0.98
mark(s(z0)) → s(z0) 0.00/0.98
mark(false) → false 0.00/0.98
mark(cons(z0, z1)) → cons(z0, z1) 0.00/0.98
mark(nil) → nil
Tuples:

A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
S tuples:none
K tuples:

MARK(inf(z0)) → c13(MARK(z0)) 0.00/0.98
MARK(eq(z0, z1)) → c12(A__EQ(z0, z1)) 0.00/0.98
MARK(take(z0, z1)) → c14(MARK(z0), MARK(z1)) 0.00/0.98
A__EQ(s(z0), s(z1)) → c1(A__EQ(z0, z1)) 0.00/0.98
MARK(length(z0)) → c15(MARK(z0))
Defined Rule Symbols:

a__eq, a__inf, a__take, a__length, mark

Defined Pair Symbols:

A__EQ, MARK

Compound Symbols:

c1, c12, c13, c14, c15

0.00/0.98
0.00/0.98

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

The set S is empty
0.00/0.98
0.00/0.98

(12) BOUNDS(O(1), O(1))

0.00/0.98
0.00/0.98
0.00/0.99 EOF