YES(O(1), O(n^2)) 15.82/4.51 YES(O(1), O(n^2)) 16.48/4.69 16.48/4.69 16.48/4.69 16.48/4.69 16.48/4.69 16.48/4.69 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 16.48/4.69 16.48/4.69 16.48/4.69
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16.48/4.69
16.48/4.69
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
16.48/4.69 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
16.48/4.69 
↳2 CdtProblem
16.48/4.69 
↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
16.48/4.69 
↳4 CdtProblem
16.48/4.69 
↳5 CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID))
16.48/4.69 
↳6 CdtProblem
16.48/4.69 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
16.48/4.69 
↳8 CdtProblem
16.48/4.69 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
16.48/4.69 
↳10 CdtProblem
16.48/4.69 
↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
16.48/4.69 
↳12 CdtProblem
16.48/4.69 
↳13 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
16.48/4.69 
↳14 CdtProblem
16.48/4.69 
↳15 SIsEmptyProof (BOTH BOUNDS(ID, ID))
16.48/4.69 
↳16 BOUNDS(O(1), O(1))
16.48/4.69
16.48/4.69 16.48/4.69
16.48/4.69
16.48/4.69

(0) Obligation:

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

minus(x, 0) → x 16.48/4.69
minus(s(x), s(y)) → minus(x, y) 16.48/4.69
quot(0, s(y)) → 0 16.48/4.69
quot(s(x), s(y)) → s(quot(minus(x, y), s(y))) 16.48/4.69
plus(0, y) → y 16.48/4.69
plus(s(x), y) → s(plus(x, y)) 16.48/4.69
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0))) 16.48/4.69
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

Rewrite Strategy: INNERMOST
16.48/4.69
16.48/4.69

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

Converted CpxTRS to CDT
16.48/4.69
16.48/4.69

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 16.48/4.69
minus(s(z0), s(z1)) → minus(z0, z1) 16.48/4.69
quot(0, s(z0)) → 0 16.48/4.69
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 16.48/4.69
plus(0, z0) → z0 16.48/4.69
plus(s(z0), z1) → s(plus(z0, z1)) 16.48/4.69
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0))) 16.48/4.69
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.48/4.69
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.48/4.69
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.48/4.69
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(PLUS(minus(z1, s(s(z2))), minus(z0, s(0))), MINUS(z1, s(s(z2))), MINUS(z0, s(0))) 16.48/4.69
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(plus(z1, s(s(z2))), plus(z0, s(0))), PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.48/4.69
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.48/4.69
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.48/4.69
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(PLUS(minus(z1, s(s(z2))), minus(z0, s(0))), MINUS(z1, s(s(z2))), MINUS(z0, s(0))) 16.48/4.69
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(plus(z1, s(s(z2))), plus(z0, s(0))), PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
K tuples:none
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6, c7

16.48/4.69
16.48/4.69

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

Removed 2 trailing tuple parts
16.48/4.69
16.48/4.69

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1) 16.83/4.71
quot(0, s(z0)) → 0 16.83/4.71
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 16.83/4.71
plus(0, z0) → z0 16.83/4.71
plus(s(z0), z1) → s(plus(z0, z1)) 16.83/4.71
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0))) 16.83/4.71
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0))) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0))) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0)))
K tuples:none
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6, c7

16.83/4.71
16.83/4.71

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

Split RHS of tuples not part of any SCC
16.83/4.71
16.83/4.71

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1) 16.83/4.71
quot(0, s(z0)) → 0 16.83/4.71
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 16.83/4.71
plus(0, z0) → z0 16.83/4.71
plus(s(z0), z1) → s(plus(z0, z1)) 16.83/4.71
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0))) 16.83/4.71
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
K tuples:none
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

16.83/4.71
16.83/4.71

(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.

PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
We considered the (Usable) Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1)
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 16.83/4.71

POL(0) = 0    16.83/4.71
POL(MINUS(x1, x2)) = 0    16.83/4.71
POL(PLUS(x1, x2)) = [3]x2    16.83/4.71
POL(QUOT(x1, x2)) = 0    16.83/4.71
POL(c(x1)) = x1    16.83/4.71
POL(c1(x1)) = x1    16.83/4.71
POL(c3(x1, x2)) = x1 + x2    16.83/4.71
POL(c5(x1)) = x1    16.83/4.71
POL(c7(x1, x2)) = x1 + x2    16.83/4.71
POL(minus(x1, x2)) = [4] + [2]x1    16.83/4.71
POL(plus(x1, x2)) = [4]x2    16.83/4.71
POL(s(x1)) = [2]   
16.83/4.71
16.83/4.71

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1) 16.83/4.71
quot(0, s(z0)) → 0 16.83/4.71
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 16.83/4.71
plus(0, z0) → z0 16.83/4.71
plus(s(z0), z1) → s(plus(z0, z1)) 16.83/4.71
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0))) 16.83/4.71
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
K tuples:

PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

16.83/4.71
16.83/4.71

(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.

PLUS(s(z0), z1) → c5(PLUS(z0, z1))
We considered the (Usable) Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1)
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 16.83/4.71

POL(0) = [2]    16.83/4.71
POL(MINUS(x1, x2)) = 0    16.83/4.71
POL(PLUS(x1, x2)) = [2]x1 + x2    16.83/4.71
POL(QUOT(x1, x2)) = 0    16.83/4.71
POL(c(x1)) = x1    16.83/4.71
POL(c1(x1)) = x1    16.83/4.71
POL(c3(x1, x2)) = x1 + x2    16.83/4.71
POL(c5(x1)) = x1    16.83/4.71
POL(c7(x1, x2)) = x1 + x2    16.83/4.71
POL(minus(x1, x2)) = 0    16.83/4.71
POL(plus(x1, x2)) = [1] + [4]x1 + x2    16.83/4.71
POL(s(x1)) = [1] + x1   
16.83/4.71
16.83/4.71

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1) 16.83/4.71
quot(0, s(z0)) → 0 16.83/4.71
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 16.83/4.71
plus(0, z0) → z0 16.83/4.71
plus(s(z0), z1) → s(plus(z0, z1)) 16.83/4.71
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0))) 16.83/4.71
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:

PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0))) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1))
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

16.83/4.71
16.83/4.71

(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.

QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1)
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 16.83/4.71

POL(0) = 0    16.83/4.71
POL(MINUS(x1, x2)) = 0    16.83/4.71
POL(PLUS(x1, x2)) = 0    16.83/4.71
POL(QUOT(x1, x2)) = [5]x1    16.83/4.71
POL(c(x1)) = x1    16.83/4.71
POL(c1(x1)) = x1    16.83/4.71
POL(c3(x1, x2)) = x1 + x2    16.83/4.71
POL(c5(x1)) = x1    16.83/4.71
POL(c7(x1, x2)) = x1 + x2    16.83/4.71
POL(minus(x1, x2)) = [3] + x1    16.83/4.71
POL(plus(x1, x2)) = 0    16.83/4.71
POL(s(x1)) = [4] + x1   
16.83/4.71
16.83/4.71

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1) 16.83/4.71
quot(0, s(z0)) → 0 16.83/4.71
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 16.83/4.71
plus(0, z0) → z0 16.83/4.71
plus(s(z0), z1) → s(plus(z0, z1)) 16.83/4.71
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0))) 16.83/4.71
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:

PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0))) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

16.83/4.71
16.83/4.71

(13) 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.

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
We considered the (Usable) Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1)
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation : 16.83/4.71

POL(0) = 0    16.83/4.71
POL(MINUS(x1, x2)) = [2]x1    16.83/4.71
POL(PLUS(x1, x2)) = [2]x1 + [2]x2    16.83/4.71
POL(QUOT(x1, x2)) = [2]x12    16.83/4.71
POL(c(x1)) = x1    16.83/4.71
POL(c1(x1)) = x1    16.83/4.71
POL(c3(x1, x2)) = x1 + x2    16.83/4.71
POL(c5(x1)) = x1    16.83/4.71
POL(c7(x1, x2)) = x1 + x2    16.83/4.71
POL(minus(x1, x2)) = x1    16.83/4.71
POL(plus(x1, x2)) = x2 + [2]x1·x2    16.83/4.71
POL(s(x1)) = [2] + x1   
16.83/4.71
16.83/4.71

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 16.83/4.71
minus(s(z0), s(z1)) → minus(z0, z1) 16.83/4.71
quot(0, s(z0)) → 0 16.83/4.71
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 16.83/4.71
plus(0, z0) → z0 16.83/4.71
plus(s(z0), z1) → s(plus(z0, z1)) 16.83/4.71
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0))) 16.83/4.71
plus(plus(z0, s(0)), plus(z1, s(s(z2)))) → plus(plus(z1, s(s(z2))), plus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
S tuples:none
K tuples:

PLUS(plus(z0, s(0)), plus(z1, s(s(z2)))) → c7(PLUS(z1, s(s(z2))), PLUS(z0, s(0))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 16.83/4.71
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0))) 16.83/4.71
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 16.83/4.71
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 16.83/4.71
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c7, c

16.83/4.71
16.83/4.71

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

The set S is empty
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16.83/4.71

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

16.83/4.71
16.83/4.71
16.83/4.74 EOF