YES(O(1), O(n^2)) 11.85/3.46 YES(O(1), O(n^2)) 11.85/3.49 11.85/3.49 11.85/3.49 11.85/3.49 11.85/3.49 11.85/3.49 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 11.85/3.49 11.85/3.49 11.85/3.49
11.85/3.49 11.85/3.49 11.85/3.49
11.85/3.49
11.85/3.49

(0) Obligation:

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

minus(x, 0) → x 11.85/3.49
minus(s(x), s(y)) → minus(x, y) 11.85/3.49
quot(0, s(y)) → 0 11.85/3.49
quot(s(x), s(y)) → s(quot(minus(x, y), s(y))) 11.85/3.49
plus(0, y) → y 11.85/3.49
plus(s(x), y) → s(plus(x, y)) 11.85/3.49
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))

Rewrite Strategy: INNERMOST
11.85/3.49
11.85/3.49

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

Converted CpxTRS to CDT
11.85/3.49
11.85/3.49

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 11.85/3.49
minus(s(z0), s(z1)) → minus(z0, z1) 11.85/3.49
quot(0, s(z0)) → 0 11.85/3.49
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 11.85/3.49
plus(0, z0) → z0 11.85/3.49
plus(s(z0), z1) → s(plus(z0, z1)) 11.85/3.49
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
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)))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
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)))
K tuples:none
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6

12.30/3.52
12.30/3.52

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

Removed 1 trailing tuple parts
12.30/3.52
12.30/3.52

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 12.30/3.52
minus(s(z0), s(z1)) → minus(z0, z1) 12.30/3.52
quot(0, s(z0)) → 0 12.30/3.52
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 12.30/3.52
plus(0, z0) → z0 12.30/3.52
plus(s(z0), z1) → s(plus(z0, z1)) 12.30/3.52
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c6(MINUS(z1, s(s(z2))), MINUS(z0, s(0)))
K tuples:none
Defined Rule Symbols:

minus, quot, plus

Defined Pair Symbols:

MINUS, QUOT, PLUS

Compound Symbols:

c1, c3, c5, c6

12.30/3.52
12.30/3.52

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

Split RHS of tuples not part of any SCC
12.30/3.52
12.30/3.52

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 12.30/3.52
minus(s(z0), s(z1)) → minus(z0, z1) 12.30/3.52
quot(0, s(z0)) → 0 12.30/3.52
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 12.30/3.52
plus(0, z0) → z0 12.30/3.52
plus(s(z0), z1) → s(plus(z0, z1)) 12.30/3.52
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
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)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
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, c

12.30/3.52
12.30/3.52

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

minus(z0, 0) → z0 12.30/3.52
minus(s(z0), s(z1)) → minus(z0, z1)
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
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 : 12.30/3.52

POL(0) = 0    12.30/3.52
POL(MINUS(x1, x2)) = 0    12.30/3.52
POL(PLUS(x1, x2)) = [3]x1 + [3]x2    12.30/3.52
POL(QUOT(x1, x2)) = 0    12.30/3.52
POL(c(x1)) = x1    12.30/3.52
POL(c1(x1)) = x1    12.30/3.52
POL(c3(x1, x2)) = x1 + x2    12.30/3.52
POL(c5(x1)) = x1    12.30/3.52
POL(minus(x1, x2)) = [2]x1    12.30/3.52
POL(s(x1)) = [1] + x1   
12.30/3.52
12.30/3.52

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 12.30/3.52
minus(s(z0), s(z1)) → minus(z0, z1) 12.30/3.52
quot(0, s(z0)) → 0 12.30/3.52
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 12.30/3.52
plus(0, z0) → z0 12.30/3.52
plus(s(z0), z1) → s(plus(z0, z1)) 12.30/3.52
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
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)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
K tuples:

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, c

12.30/3.52
12.30/3.52

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

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

PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
12.30/3.52
12.30/3.52

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 12.30/3.52
minus(s(z0), s(z1)) → minus(z0, z1) 12.30/3.52
quot(0, s(z0)) → 0 12.30/3.52
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 12.30/3.52
plus(0, z0) → z0 12.30/3.52
plus(s(z0), z1) → s(plus(z0, z1)) 12.30/3.52
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
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)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:

PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
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, c

12.30/3.52
12.30/3.52

(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 12.30/3.52
minus(s(z0), s(z1)) → minus(z0, z1)
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
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 : 12.30/3.52

POL(0) = [1]    12.30/3.52
POL(MINUS(x1, x2)) = 0    12.30/3.52
POL(PLUS(x1, x2)) = 0    12.30/3.52
POL(QUOT(x1, x2)) = x1    12.30/3.52
POL(c(x1)) = x1    12.30/3.52
POL(c1(x1)) = x1    12.30/3.52
POL(c3(x1, x2)) = x1 + x2    12.30/3.52
POL(c5(x1)) = x1    12.30/3.52
POL(minus(x1, x2)) = x1    12.30/3.52
POL(s(x1)) = [1] + x1   
12.30/3.52
12.30/3.52

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 12.30/3.52
minus(s(z0), s(z1)) → minus(z0, z1) 12.30/3.52
quot(0, s(z0)) → 0 12.30/3.52
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 12.30/3.52
plus(0, z0) → z0 12.30/3.52
plus(s(z0), z1) → s(plus(z0, z1)) 12.30/3.52
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
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(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0))) 12.30/3.52
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, c

12.30/3.52
12.30/3.52

(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 12.30/3.52
minus(s(z0), s(z1)) → minus(z0, z1)
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
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 : 12.30/3.52

POL(0) = 0    12.30/3.52
POL(MINUS(x1, x2)) = x1    12.30/3.52
POL(PLUS(x1, x2)) = x1 + [2]x2    12.30/3.52
POL(QUOT(x1, x2)) = [2]x12    12.30/3.52
POL(c(x1)) = x1    12.30/3.52
POL(c1(x1)) = x1    12.30/3.52
POL(c3(x1, x2)) = x1 + x2    12.30/3.52
POL(c5(x1)) = x1    12.30/3.52
POL(minus(x1, x2)) = x1    12.30/3.52
POL(s(x1)) = [2] + x1   
12.30/3.52
12.30/3.52

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0 12.30/3.52
minus(s(z0), s(z1)) → minus(z0, z1) 12.30/3.52
quot(0, s(z0)) → 0 12.30/3.52
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1))) 12.30/3.52
plus(0, z0) → z0 12.30/3.52
plus(s(z0), z1) → s(plus(z0, z1)) 12.30/3.52
plus(minus(z0, s(0)), minus(z1, s(s(z2)))) → plus(minus(z1, s(s(z2))), minus(z0, s(0)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1)) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0)))
S tuples:none
K tuples:

PLUS(s(z0), z1) → c5(PLUS(z0, z1)) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z1, s(s(z2)))) 12.30/3.52
PLUS(minus(z0, s(0)), minus(z1, s(s(z2)))) → c(MINUS(z0, s(0))) 12.30/3.52
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1)) 12.30/3.52
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, c

12.30/3.52
12.30/3.52

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

The set S is empty
12.30/3.52
12.30/3.52

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

12.30/3.52
12.30/3.52
12.30/3.57 EOF