YES(O(1), O(n^2)) 5.90/1.90 YES(O(1), O(n^2)) 5.90/1.93 5.90/1.93 5.90/1.93 5.90/1.93 5.90/1.93 5.90/1.93 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 5.90/1.93 5.90/1.93 5.90/1.93
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
5.90/1.93 
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
5.90/1.93 
↳2 CdtProblem
5.90/1.93 
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
5.90/1.93 
↳4 CdtProblem
5.90/1.93 
↳5 CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
5.90/1.93 
↳6 CdtProblem
5.90/1.93 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
5.90/1.93 
↳8 CdtProblem
5.90/1.93 
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
5.90/1.93 
↳10 CdtProblem
5.90/1.93 
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID))
5.90/1.93 
↳12 BOUNDS(O(1), O(1))
5.90/1.93
5.90/1.93 5.90/1.93
5.90/1.93
5.90/1.93

(0) Obligation:

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

le(0, y) → true 5.90/1.93
le(s(x), 0) → false 5.90/1.93
le(s(x), s(y)) → le(x, y) 5.90/1.93
minus(x, 0) → x 5.90/1.93
minus(s(x), s(y)) → minus(x, y) 5.90/1.93
mod(0, y) → 0 5.90/1.93
mod(s(x), 0) → 0 5.90/1.93
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y)) 5.90/1.93
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y)) 5.90/1.93
if_mod(false, s(x), s(y)) → s(x)

Rewrite Strategy: INNERMOST
5.90/1.93
5.90/1.93

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

Converted CpxTRS to CDT
5.90/1.93
5.90/1.93

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 5.90/1.93
le(s(z0), 0) → false 5.90/1.93
le(s(z0), s(z1)) → le(z0, z1) 5.90/1.93
minus(z0, 0) → z0 5.90/1.93
minus(s(z0), s(z1)) → minus(z0, z1) 5.90/1.93
mod(0, z0) → 0 5.90/1.93
mod(s(z0), 0) → 0 5.90/1.93
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 5.90/1.93
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 5.90/1.93
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.94
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.94
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.94
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.94
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.94
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.94
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

le, minus, mod, if_mod

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

5.90/1.94
5.90/1.94

(3) 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.

IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

minus(z0, 0) → z0 5.90/1.94
minus(s(z0), s(z1)) → minus(z0, z1) 5.90/1.94
le(0, z0) → true 5.90/1.94
le(s(z0), 0) → false 5.90/1.94
le(s(z0), s(z1)) → le(z0, z1)
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.94
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.94
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.94
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.90/1.94

POL(0) = [1]    5.90/1.94
POL(IF_MOD(x1, x2, x3)) = [2]x2    5.90/1.94
POL(LE(x1, x2)) = 0    5.90/1.95
POL(MINUS(x1, x2)) = 0    5.90/1.95
POL(MOD(x1, x2)) = [2]x1    5.90/1.95
POL(c2(x1)) = x1    5.90/1.95
POL(c4(x1)) = x1    5.90/1.95
POL(c7(x1, x2)) = x1 + x2    5.90/1.95
POL(c8(x1, x2)) = x1 + x2    5.90/1.95
POL(false) = [3]    5.90/1.95
POL(le(x1, x2)) = 0    5.90/1.95
POL(minus(x1, x2)) = x1    5.90/1.95
POL(s(x1)) = [2] + x1    5.90/1.95
POL(true) = 0   
5.90/1.95
5.90/1.95

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 5.90/1.95
le(s(z0), 0) → false 5.90/1.95
le(s(z0), s(z1)) → le(z0, z1) 5.90/1.95
minus(z0, 0) → z0 5.90/1.95
minus(s(z0), s(z1)) → minus(z0, z1) 5.90/1.95
mod(0, z0) → 0 5.90/1.95
mod(s(z0), 0) → 0 5.90/1.95
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 5.90/1.95
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 5.90/1.95
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.95
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
K tuples:

IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

le, minus, mod, if_mod

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

5.90/1.95
5.90/1.95

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

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

MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.95
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
5.90/1.95
5.90/1.95

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 5.90/1.95
le(s(z0), 0) → false 5.90/1.95
le(s(z0), s(z1)) → le(z0, z1) 5.90/1.95
minus(z0, 0) → z0 5.90/1.95
minus(s(z0), s(z1)) → minus(z0, z1) 5.90/1.95
mod(0, z0) → 0 5.90/1.95
mod(s(z0), 0) → 0 5.90/1.95
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 5.90/1.95
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 5.90/1.95
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.95
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
K tuples:

IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
Defined Rule Symbols:

le, minus, mod, if_mod

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

5.90/1.95
5.90/1.95

(7) 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)) → c4(MINUS(z0, z1))
We considered the (Usable) Rules:

minus(z0, 0) → z0 5.90/1.95
minus(s(z0), s(z1)) → minus(z0, z1) 5.90/1.95
le(0, z0) → true 5.90/1.95
le(s(z0), 0) → false 5.90/1.95
le(s(z0), s(z1)) → le(z0, z1)
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.95
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.90/1.95

POL(0) = 0    5.90/1.95
POL(IF_MOD(x1, x2, x3)) = x2 + x12 + x22    5.90/1.95
POL(LE(x1, x2)) = 0    5.90/1.95
POL(MINUS(x1, x2)) = [1] + x1    5.90/1.95
POL(MOD(x1, x2)) = [2] + [2]x1 + x12    5.90/1.95
POL(c2(x1)) = x1    5.90/1.95
POL(c4(x1)) = x1    5.90/1.95
POL(c7(x1, x2)) = x1 + x2    5.90/1.95
POL(c8(x1, x2)) = x1 + x2    5.90/1.95
POL(false) = [1]    5.90/1.95
POL(le(x1, x2)) = [2]    5.90/1.95
POL(minus(x1, x2)) = [1] + x1    5.90/1.95
POL(s(x1)) = [2] + x1    5.90/1.95
POL(true) = 0   
5.90/1.95
5.90/1.95

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 5.90/1.95
le(s(z0), 0) → false 5.90/1.95
le(s(z0), s(z1)) → le(z0, z1) 5.90/1.95
minus(z0, 0) → z0 5.90/1.95
minus(s(z0), s(z1)) → minus(z0, z1) 5.90/1.95
mod(0, z0) → 0 5.90/1.95
mod(s(z0), 0) → 0 5.90/1.95
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 5.90/1.95
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 5.90/1.95
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.95
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1))
K tuples:

IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1))
Defined Rule Symbols:

le, minus, mod, if_mod

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

5.90/1.95
5.90/1.95

(9) 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.

LE(s(z0), s(z1)) → c2(LE(z0, z1))
We considered the (Usable) Rules:

minus(z0, 0) → z0 5.90/1.95
minus(s(z0), s(z1)) → minus(z0, z1) 5.90/1.95
le(0, z0) → true 5.90/1.95
le(s(z0), 0) → false 5.90/1.95
le(s(z0), s(z1)) → le(z0, z1)
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.95
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.90/1.95

POL(0) = 0    5.90/1.95
POL(IF_MOD(x1, x2, x3)) = x2·x3 + [3]x22    5.90/1.95
POL(LE(x1, x2)) = x1    5.90/1.95
POL(MINUS(x1, x2)) = [2]x1    5.90/1.95
POL(MOD(x1, x2)) = x2 + x1·x2 + [3]x12    5.90/1.95
POL(c2(x1)) = x1    5.90/1.95
POL(c4(x1)) = x1    5.90/1.95
POL(c7(x1, x2)) = x1 + x2    5.90/1.95
POL(c8(x1, x2)) = x1 + x2    5.90/1.95
POL(false) = [3]    5.90/1.95
POL(le(x1, x2)) = 0    5.90/1.95
POL(minus(x1, x2)) = x1    5.90/1.95
POL(s(x1)) = [2] + x1    5.90/1.95
POL(true) = 0   
5.90/1.95
5.90/1.95

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 5.90/1.95
le(s(z0), 0) → false 5.90/1.95
le(s(z0), s(z1)) → le(z0, z1) 5.90/1.95
minus(z0, 0) → z0 5.90/1.95
minus(s(z0), s(z1)) → minus(z0, z1) 5.90/1.95
mod(0, z0) → 0 5.90/1.95
mod(s(z0), 0) → 0 5.90/1.95
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 5.90/1.95
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 5.90/1.95
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.95
IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:none
K tuples:

IF_MOD(true, s(z0), s(z1)) → c8(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 5.90/1.95
MOD(s(z0), s(z1)) → c7(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 5.90/1.95
MINUS(s(z0), s(z1)) → c4(MINUS(z0, z1)) 5.90/1.95
LE(s(z0), s(z1)) → c2(LE(z0, z1))
Defined Rule Symbols:

le, minus, mod, if_mod

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c4, c7, c8

5.90/1.95
5.90/1.95

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

The set S is empty
5.90/1.95
5.90/1.95

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

5.90/1.95
5.90/1.95
5.90/2.00 EOF