The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 93 ms)
↳8 CdtProblem
↳9 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳10 CdtProblem
↳11 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 CdtProblem
↳13 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳14 CdtProblem
↳15 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳16 CdtProblem
↳17 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳18 CdtProblem
↳19 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 25 ms)
↳20 CdtProblem
↳21 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 0 ms)
↳22 CdtProblem
↳23 CdtNarrowingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳24 CdtProblem
↳25 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳26 CdtProblem
↳27 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 26 ms)
↳28 CdtProblem
↳29 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 36 ms)
↳30 CdtProblem
↳31 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 21 ms)
↳32 CdtProblem
↳33 CdtForwardInstantiationProof (BOTH BOUNDS(ID, ID), 0 ms)
↳34 CdtProblem
↳35 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳36 CdtProblem
↳37 CdtRewritingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳38 CdtProblem
↳39 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 3 ms)
↳40 CdtProblem
↳41 CdtRewritingProof (BOTH BOUNDS(ID, ID), 0 ms)
↳42 CdtProblem
↳43 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 0 ms)
↳44 CdtProblem
↳45 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))), 10 ms)
↳46 CdtProblem
↳47 SIsEmptyProof (⇔, 0 ms)
↳48 BOUNDS(O(1), O(1))

(0) Obligation:

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

cond(true, x) → cond(odd(x), p(x))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

COND(true, z0) → c(COND(odd(z0), p(z0)), ODD(z0), P(z0))
ODD(s(s(z0))) → c3(ODD(z0))
S tuples:

COND(true, z0) → c(COND(odd(z0), p(z0)), ODD(z0), P(z0))
ODD(s(s(z0))) → c3(ODD(z0))
K tuples:none
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

COND, ODD

Compound Symbols:

c, c3

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

Use narrowing to replace COND(true, z0) → c(COND(odd(z0), p(z0)), ODD(z0), P(z0)) by

COND(true, 0) → c(COND(odd(0), 0), ODD(0), P(0))
COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
COND(true, 0) → c(COND(false, p(0)), ODD(0), P(0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, x0) → c

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, 0) → c(COND(odd(0), 0), ODD(0), P(0))
COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
COND(true, 0) → c(COND(false, p(0)), ODD(0), P(0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, x0) → c
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, 0) → c(COND(odd(0), 0), ODD(0), P(0))
COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
COND(true, 0) → c(COND(false, p(0)), ODD(0), P(0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, x0) → c
K tuples:none
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

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

Removed 2 trailing nodes:

COND(true, x0) → c
COND(true, 0) → c(COND(false, p(0)), ODD(0), P(0))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, 0) → c(COND(odd(0), 0), ODD(0), P(0))
COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, 0) → c(COND(odd(0), 0), ODD(0), P(0))
COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
K tuples:none
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c

(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.

COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
We considered the (Usable) Rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(s(z0)) → z0
And the Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, 0) → c(COND(odd(0), 0), ODD(0), P(0))
COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(COND(x1, x2)) = [4]x2   
POL(ODD(x1)) = 0   
POL(P(x1)) = 0   
POL(c(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(false) = [3]   
POL(odd(x1)) = 0   
POL(p(x1)) = x1   
POL(s(x1)) = [1] + x1   
POL(true) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, 0) → c(COND(odd(0), 0), ODD(0), P(0))
COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, 0) → c(COND(odd(0), 0), ODD(0), P(0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
K tuples:

COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c

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

Use narrowing to replace COND(true, 0) → c(COND(odd(0), 0), ODD(0), P(0)) by

COND(true, 0) → c(COND(false, 0), ODD(0), P(0))
COND(true, 0) → c

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, 0) → c(COND(false, 0), ODD(0), P(0))
COND(true, 0) → c
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, 0) → c(COND(false, 0), ODD(0), P(0))
COND(true, 0) → c
K tuples:

COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

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

Removed 2 trailing nodes:

COND(true, 0) → c
COND(true, 0) → c(COND(false, 0), ODD(0), P(0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
K tuples:

COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c

(13) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0))) by

COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(x0)) → c

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(x0)) → c
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
K tuples:

COND(true, s(z0)) → c(COND(odd(s(z0)), z0), ODD(s(z0)), P(s(z0)))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

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

Removed 2 trailing nodes:

COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(x0)) → c

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0)))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
K tuples:none
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c

(17) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace COND(true, s(0)) → c(COND(true, p(s(0))), ODD(s(0)), P(s(0))) by

COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(0)) → c

(18) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(0)) → c
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(0)) → c
K tuples:none
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

(19) 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.

COND(true, s(0)) → c
We considered the (Usable) Rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(s(z0)) → z0
And the Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(0)) → c
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [5]   
POL(COND(x1, x2)) = [2]   
POL(ODD(x1)) = 0   
POL(P(x1)) = 0   
POL(c) = 0   
POL(c(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(false) = [3]   
POL(odd(x1)) = [3] + [3]x1   
POL(p(x1)) = [2] + [3]x1   
POL(s(x1)) = 0   
POL(true) = [3]   

(20) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(0)) → c
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
K tuples:

COND(true, s(0)) → c
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

(21) 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.

COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
We considered the (Usable) Rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(s(z0)) → z0
And the Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(0)) → c
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [4]   
POL(COND(x1, x2)) = [4]x2   
POL(ODD(x1)) = 0   
POL(P(x1)) = 0   
POL(c) = 0   
POL(c(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(false) = [3]   
POL(odd(x1)) = 0   
POL(p(x1)) = x1   
POL(s(x1)) = [4] + x1   
POL(true) = 0   

(22) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(0)) → c
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0))))
K tuples:

COND(true, s(0)) → c
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

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

Use narrowing to replace COND(true, s(s(z0))) → c(COND(odd(z0), p(s(s(z0)))), ODD(s(s(z0))), P(s(s(z0)))) by

COND(true, s(s(x0))) → c(COND(odd(x0), s(x0)), ODD(s(s(x0))), P(s(s(x0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))

(24) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(0)) → c
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(x0))) → c(COND(odd(x0), s(x0)), ODD(s(s(x0))), P(s(s(x0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
K tuples:

COND(true, s(0)) → c
COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c, c

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

Removed 2 trailing nodes:

COND(true, s(0)) → c(COND(true, 0), ODD(s(0)), P(s(0)))
COND(true, s(0)) → c

(26) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(x0))) → c(COND(odd(x0), s(x0)), ODD(s(s(x0))), P(s(s(x0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
K tuples:none
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

(27) 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.

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
We considered the (Usable) Rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(s(z0)) → z0
And the Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [5]   
POL(COND(x1, x2)) = [4]   
POL(ODD(x1)) = 0   
POL(P(x1)) = 0   
POL(c(x1)) = x1   
POL(c(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(false) = [1]   
POL(odd(x1)) = [3] + [3]x1   
POL(p(x1)) = [4]   
POL(s(x1)) = [2]   
POL(true) = [2]   

(28) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(x0))) → c(COND(odd(x0), s(x0)), ODD(s(s(x0))), P(s(s(x0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

(29) 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.

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
We considered the (Usable) Rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(s(z0)) → z0
And the Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(COND(x1, x2)) = x2   
POL(ODD(x1)) = 0   
POL(P(x1)) = 0   
POL(c(x1)) = x1   
POL(c(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(false) = [3]   
POL(odd(x1)) = [4] + [3]x1   
POL(p(x1)) = x1   
POL(s(x1)) = [1] + x1   
POL(true) = [3]   

(30) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

(31) 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.

COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
We considered the (Usable) Rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(s(z0)) → z0
And the Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [4]   
POL(COND(x1, x2)) = [3] + [4]x1   
POL(ODD(x1)) = 0   
POL(P(x1)) = 0   
POL(c(x1)) = x1   
POL(c(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(false) = 0   
POL(odd(x1)) = [2]   
POL(p(x1)) = 0   
POL(s(x1)) = 0   
POL(true) = [2]   

(32) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
S tuples:

ODD(s(s(z0))) → c3(ODD(z0))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

ODD, COND

Compound Symbols:

c3, c, c

(33) CdtForwardInstantiationProof (BOTH BOUNDS(ID, ID) transformation)

Use forward instantiation to replace ODD(s(s(z0))) → c3(ODD(z0)) by

ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))

(34) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
S tuples:

COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

COND, ODD

Compound Symbols:

c, c, c3

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

Removed 1 trailing nodes:

COND(true, s(s(0))) → c(COND(false, p(s(s(0)))), ODD(s(s(0))), P(s(s(0))))

(36) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
S tuples:

COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

COND, ODD

Compound Symbols:

c, c, c3

(37) CdtRewritingProof (BOTH BOUNDS(ID, ID) transformation)

Used rewriting to replace COND(true, s(s(s(0)))) → c(COND(true, p(s(s(s(0))))), ODD(s(s(s(0)))), P(s(s(s(0))))) by COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))

(38) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
S tuples:

COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

COND, ODD

Compound Symbols:

c, c, c3

(39) 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.

COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
We considered the (Usable) Rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(s(z0)) → z0
And the Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [4]   
POL(COND(x1, x2)) = [4]x2   
POL(ODD(x1)) = 0   
POL(P(x1)) = 0   
POL(c(x1)) = x1   
POL(c(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(false) = [5]   
POL(odd(x1)) = 0   
POL(p(x1)) = x1   
POL(s(x1)) = [4] + x1   
POL(true) = 0   

(40) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
S tuples:

COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

COND, ODD

Compound Symbols:

c, c, c3

(41) CdtRewritingProof (BOTH BOUNDS(ID, ID) transformation)

Used rewriting to replace COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), p(s(s(s(s(z0)))))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0)))))) by COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))

(42) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
S tuples:

ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

COND, ODD

Compound Symbols:

c, c, c3

(43) 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.

COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
We considered the (Usable) Rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
And the Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(COND(x1, x2)) = [1] + x2   
POL(ODD(x1)) = 0   
POL(P(x1)) = 0   
POL(c(x1)) = x1   
POL(c(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(false) = [4]   
POL(odd(x1)) = [4]x1   
POL(s(x1)) = [1] + x1   
POL(true) = 0   

(44) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
S tuples:

ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

COND, ODD

Compound Symbols:

c, c, c3

(45) 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.

ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
We considered the (Usable) Rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
And the Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [2]   
POL(COND(x1, x2)) = [1] + x22   
POL(ODD(x1)) = x1   
POL(P(x1)) = [1]   
POL(c(x1)) = x1   
POL(c(x1, x2, x3)) = x1 + x2 + x3   
POL(c3(x1)) = x1   
POL(false) = 0   
POL(odd(x1)) = 0   
POL(s(x1)) = [1] + x1   
POL(true) = 0   

(46) Obligation:

Complexity Dependency Tuples Problem
Rules:

cond(true, z0) → cond(odd(z0), p(z0))
odd(0) → false
odd(s(0)) → true
odd(s(s(z0))) → odd(z0)
p(0) → 0
p(s(z0)) → z0
Tuples:

COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(x0))) → c(ODD(s(s(x0))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
S tuples:none
K tuples:

COND(true, s(s(x0))) → c(ODD(s(s(x0))))
COND(true, s(s(z0))) → c(COND(odd(z0), s(z0)), ODD(s(s(z0))), P(s(s(z0))))
COND(true, s(s(s(0)))) → c(COND(true, s(s(0))), ODD(s(s(s(0)))), P(s(s(s(0)))))
COND(true, s(s(s(s(z0))))) → c(COND(odd(z0), s(s(s(z0)))), ODD(s(s(s(s(z0))))), P(s(s(s(s(z0))))))
ODD(s(s(s(s(y0))))) → c3(ODD(s(s(y0))))
Defined Rule Symbols:

cond, odd, p

Defined Pair Symbols:

COND, ODD

Compound Symbols:

c, c, c3

(47) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(48) BOUNDS(O(1), O(1))