YES(O(1), O(n^2)) 5.88/1.95 YES(O(1), O(n^2)) 6.31/2.00 6.31/2.00 6.31/2.00 6.31/2.00 6.31/2.00 6.31/2.00 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 6.31/2.00 6.31/2.00 6.31/2.00
6.31/2.00 6.31/2.00 6.31/2.00
6.31/2.00
6.31/2.00

(0) Obligation:

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

le(0, y) → true 6.31/2.00
le(s(x), 0) → false 6.31/2.00
le(s(x), s(y)) → le(x, y) 6.31/2.00
pred(s(x)) → x 6.31/2.00
minus(x, 0) → x 6.31/2.00
minus(x, s(y)) → pred(minus(x, y)) 6.31/2.00
mod(0, y) → 0 6.31/2.00
mod(s(x), 0) → 0 6.31/2.00
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y)) 6.31/2.00
if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y)) 6.31/2.00
if_mod(false, s(x), s(y)) → s(x)

Rewrite Strategy: INNERMOST
6.31/2.00
6.31/2.00

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

Converted CpxTRS to CDT
6.31/2.00
6.31/2.00

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 6.31/2.00
le(s(z0), 0) → false 6.31/2.00
le(s(z0), s(z1)) → le(z0, z1) 6.31/2.00
pred(s(z0)) → z0 6.31/2.00
minus(z0, 0) → z0 6.31/2.00
minus(z0, s(z1)) → pred(minus(z0, z1)) 6.31/2.00
mod(0, z0) → 0 6.31/2.00
mod(s(z0), 0) → 0 6.31/2.00
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 6.31/2.00
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 6.31/2.00
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.00
MINUS(z0, s(z1)) → c5(PRED(minus(z0, z1)), MINUS(z0, z1)) 6.31/2.00
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.00
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.00
MINUS(z0, s(z1)) → c5(PRED(minus(z0, z1)), MINUS(z0, z1)) 6.31/2.00
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.00
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

le, pred, minus, mod, if_mod

Defined Pair Symbols:

LE, MINUS, MOD, IF_MOD

Compound Symbols:

c2, c5, c8, c9

6.31/2.00
6.31/2.00

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

Removed 1 trailing tuple parts
6.31/2.00
6.31/2.00

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 6.31/2.00
le(s(z0), 0) → false 6.31/2.00
le(s(z0), s(z1)) → le(z0, z1) 6.31/2.00
pred(s(z0)) → z0 6.31/2.00
minus(z0, 0) → z0 6.31/2.00
minus(z0, s(z1)) → pred(minus(z0, z1)) 6.31/2.00
mod(0, z0) → 0 6.31/2.00
mod(s(z0), 0) → 0 6.31/2.00
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 6.31/2.00
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 6.31/2.00
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.00
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.01
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.01
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.01
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.01
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.01
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

le, pred, minus, mod, if_mod

Defined Pair Symbols:

LE, MOD, IF_MOD, MINUS

Compound Symbols:

c2, c8, c9, c5

6.31/2.01
6.31/2.01

(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.

IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

minus(z0, 0) → z0 6.31/2.01
minus(z0, s(z1)) → pred(minus(z0, z1)) 6.31/2.01
pred(s(z0)) → z0 6.31/2.01
le(0, z0) → true 6.31/2.01
le(s(z0), 0) → false 6.31/2.01
le(s(z0), s(z1)) → le(z0, z1)
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.01
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.01
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.01
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 6.31/2.01

POL(0) = [1]    6.31/2.01
POL(IF_MOD(x1, x2, x3)) = [4]x2    6.31/2.01
POL(LE(x1, x2)) = 0    6.31/2.01
POL(MINUS(x1, x2)) = 0    6.31/2.01
POL(MOD(x1, x2)) = [4]x1    6.31/2.02
POL(c2(x1)) = x1    6.31/2.02
POL(c5(x1)) = x1    6.31/2.02
POL(c8(x1, x2)) = x1 + x2    6.31/2.02
POL(c9(x1, x2)) = x1 + x2    6.31/2.02
POL(false) = [3]    6.31/2.02
POL(le(x1, x2)) = 0    6.31/2.02
POL(minus(x1, x2)) = x1    6.31/2.02
POL(pred(x1)) = x1    6.31/2.02
POL(s(x1)) = [2] + x1    6.31/2.02
POL(true) = 0   
6.31/2.02
6.31/2.02

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 6.31/2.02
le(s(z0), 0) → false 6.31/2.02
le(s(z0), s(z1)) → le(z0, z1) 6.31/2.02
pred(s(z0)) → z0 6.31/2.02
minus(z0, 0) → z0 6.31/2.02
minus(z0, s(z1)) → pred(minus(z0, z1)) 6.31/2.02
mod(0, z0) → 0 6.31/2.02
mod(s(z0), 0) → 0 6.31/2.02
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 6.31/2.02
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 6.31/2.02
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.02
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
K tuples:

IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

le, pred, minus, mod, if_mod

Defined Pair Symbols:

LE, MOD, IF_MOD, MINUS

Compound Symbols:

c2, c8, c9, c5

6.31/2.02
6.31/2.02

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

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

MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1))
6.31/2.02
6.31/2.02

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 6.31/2.02
le(s(z0), 0) → false 6.31/2.02
le(s(z0), s(z1)) → le(z0, z1) 6.31/2.02
pred(s(z0)) → z0 6.31/2.02
minus(z0, 0) → z0 6.31/2.02
minus(z0, s(z1)) → pred(minus(z0, z1)) 6.31/2.02
mod(0, z0) → 0 6.31/2.02
mod(s(z0), 0) → 0 6.31/2.02
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 6.31/2.02
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 6.31/2.02
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.02
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
S tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.02
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
K tuples:

IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0))
Defined Rule Symbols:

le, pred, minus, mod, if_mod

Defined Pair Symbols:

LE, MOD, IF_MOD, MINUS

Compound Symbols:

c2, c8, c9, c5

6.31/2.02
6.31/2.02

(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 6.31/2.02
minus(z0, s(z1)) → pred(minus(z0, z1)) 6.31/2.02
pred(s(z0)) → z0 6.31/2.02
le(0, z0) → true 6.31/2.02
le(s(z0), 0) → false 6.31/2.02
le(s(z0), s(z1)) → le(z0, z1)
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.02
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 6.31/2.02

POL(0) = 0    6.31/2.02
POL(IF_MOD(x1, x2, x3)) = [2]x22    6.31/2.02
POL(LE(x1, x2)) = x2    6.31/2.02
POL(MINUS(x1, x2)) = [3]x1    6.31/2.02
POL(MOD(x1, x2)) = [1] + x1 + [2]x12    6.31/2.02
POL(c2(x1)) = x1    6.31/2.02
POL(c5(x1)) = x1    6.31/2.02
POL(c8(x1, x2)) = x1 + x2    6.31/2.02
POL(c9(x1, x2)) = x1 + x2    6.31/2.02
POL(false) = [3]    6.31/2.02
POL(le(x1, x2)) = 0    6.31/2.02
POL(minus(x1, x2)) = x1    6.31/2.02
POL(pred(x1)) = x1    6.31/2.02
POL(s(x1)) = [1] + x1    6.31/2.02
POL(true) = 0   
6.31/2.02
6.31/2.02

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 6.31/2.02
le(s(z0), 0) → false 6.31/2.02
le(s(z0), s(z1)) → le(z0, z1) 6.31/2.02
pred(s(z0)) → z0 6.31/2.02
minus(z0, 0) → z0 6.31/2.02
minus(z0, s(z1)) → pred(minus(z0, z1)) 6.31/2.02
mod(0, z0) → 0 6.31/2.02
mod(s(z0), 0) → 0 6.31/2.02
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 6.31/2.02
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 6.31/2.02
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.02
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
S tuples:

MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
K tuples:

IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
LE(s(z0), s(z1)) → c2(LE(z0, z1))
Defined Rule Symbols:

le, pred, minus, mod, if_mod

Defined Pair Symbols:

LE, MOD, IF_MOD, MINUS

Compound Symbols:

c2, c8, c9, c5

6.31/2.02
6.31/2.02

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

minus(z0, 0) → z0 6.31/2.02
minus(z0, s(z1)) → pred(minus(z0, z1)) 6.31/2.02
pred(s(z0)) → z0 6.31/2.02
le(0, z0) → true 6.31/2.02
le(s(z0), 0) → false 6.31/2.02
le(s(z0), s(z1)) → le(z0, z1)
And the Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.02
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 6.31/2.02

POL(0) = 0    6.31/2.02
POL(IF_MOD(x1, x2, x3)) = [2]x2·x3    6.31/2.02
POL(LE(x1, x2)) = 0    6.31/2.02
POL(MINUS(x1, x2)) = x2    6.31/2.02
POL(MOD(x1, x2)) = [2]x1·x2    6.31/2.02
POL(c2(x1)) = x1    6.31/2.02
POL(c5(x1)) = x1    6.31/2.02
POL(c8(x1, x2)) = x1 + x2    6.31/2.02
POL(c9(x1, x2)) = x1 + x2    6.31/2.02
POL(false) = [3]    6.31/2.02
POL(le(x1, x2)) = 0    6.31/2.02
POL(minus(x1, x2)) = x1    6.31/2.02
POL(pred(x1)) = x1    6.31/2.02
POL(s(x1)) = [2] + x1    6.31/2.02
POL(true) = 0   
6.31/2.02
6.31/2.02

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

le(0, z0) → true 6.31/2.02
le(s(z0), 0) → false 6.31/2.02
le(s(z0), s(z1)) → le(z0, z1) 6.31/2.02
pred(s(z0)) → z0 6.31/2.02
minus(z0, 0) → z0 6.31/2.02
minus(z0, s(z1)) → pred(minus(z0, z1)) 6.31/2.02
mod(0, z0) → 0 6.31/2.02
mod(s(z0), 0) → 0 6.31/2.02
mod(s(z0), s(z1)) → if_mod(le(z1, z0), s(z0), s(z1)) 6.31/2.02
if_mod(true, s(z0), s(z1)) → mod(minus(z0, z1), s(z1)) 6.31/2.02
if_mod(false, s(z0), s(z1)) → s(z0)
Tuples:

LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.02
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
S tuples:none
K tuples:

IF_MOD(true, s(z0), s(z1)) → c9(MOD(minus(z0, z1), s(z1)), MINUS(z0, z1)) 6.31/2.02
MOD(s(z0), s(z1)) → c8(IF_MOD(le(z1, z0), s(z0), s(z1)), LE(z1, z0)) 6.31/2.02
LE(s(z0), s(z1)) → c2(LE(z0, z1)) 6.31/2.02
MINUS(z0, s(z1)) → c5(MINUS(z0, z1))
Defined Rule Symbols:

le, pred, minus, mod, if_mod

Defined Pair Symbols:

LE, MOD, IF_MOD, MINUS

Compound Symbols:

c2, c8, c9, c5

6.31/2.02
6.31/2.02

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

The set S is empty
6.31/2.02
6.31/2.02

(14) BOUNDS(O(1), O(1))

6.31/2.02
6.31/2.02
6.31/2.05 EOF