(0) Obligation:
The Runtime Complexity (innermost) of the given
CpxTRS could be proven to be
BOUNDS(1, n^2).
The TRS R consists of the following rules:
cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
cond1(true, z0) → cond2(even(z0), z0)
cond2(true, z0) → cond1(neq(z0, 0), div2(z0))
cond2(false, z0) → cond1(neq(z0, 0), p(z0))
neq(0, 0) → false
neq(0, s(z0)) → true
neq(s(z0), 0) → true
neq(s(z0), s(y)) → neq(z0, y)
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
COND1(true, z0) → c(COND2(even(z0), z0), EVEN(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), NEQ(z0, 0), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)), NEQ(z0, 0), P(z0))
NEQ(0, 0) → c3
NEQ(0, s(z0)) → c4
NEQ(s(z0), 0) → c5
NEQ(s(z0), s(y)) → c6(NEQ(z0, y))
EVEN(0) → c7
EVEN(s(0)) → c8
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(0) → c10
DIV2(s(0)) → c11
DIV2(s(s(z0))) → c12(DIV2(z0))
P(0) → c13
P(s(z0)) → c14
S tuples:
COND1(true, z0) → c(COND2(even(z0), z0), EVEN(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), NEQ(z0, 0), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)), NEQ(z0, 0), P(z0))
NEQ(0, 0) → c3
NEQ(0, s(z0)) → c4
NEQ(s(z0), 0) → c5
NEQ(s(z0), s(y)) → c6(NEQ(z0, y))
EVEN(0) → c7
EVEN(s(0)) → c8
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(0) → c10
DIV2(s(0)) → c11
DIV2(s(s(z0))) → c12(DIV2(z0))
P(0) → c13
P(s(z0)) → c14
K tuples:none
Defined Rule Symbols:
cond1, cond2, neq, even, div2, p
Defined Pair Symbols:
COND1, COND2, NEQ, EVEN, DIV2, P
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12, c13, c14
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 10 trailing nodes:
NEQ(0, s(z0)) → c4
EVEN(s(0)) → c8
NEQ(0, 0) → c3
EVEN(0) → c7
P(s(z0)) → c14
DIV2(s(0)) → c11
DIV2(0) → c10
NEQ(s(z0), s(y)) → c6(NEQ(z0, y))
P(0) → c13
NEQ(s(z0), 0) → c5
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
cond1(true, z0) → cond2(even(z0), z0)
cond2(true, z0) → cond1(neq(z0, 0), div2(z0))
cond2(false, z0) → cond1(neq(z0, 0), p(z0))
neq(0, 0) → false
neq(0, s(z0)) → true
neq(s(z0), 0) → true
neq(s(z0), s(y)) → neq(z0, y)
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
COND1(true, z0) → c(COND2(even(z0), z0), EVEN(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), NEQ(z0, 0), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)), NEQ(z0, 0), P(z0))
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
S tuples:
COND1(true, z0) → c(COND2(even(z0), z0), EVEN(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), NEQ(z0, 0), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)), NEQ(z0, 0), P(z0))
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
K tuples:none
Defined Rule Symbols:
cond1, cond2, neq, even, div2, p
Defined Pair Symbols:
COND1, COND2, EVEN, DIV2
Compound Symbols:
c, c1, c2, c9, c12
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
cond1(true, z0) → cond2(even(z0), z0)
cond2(true, z0) → cond1(neq(z0, 0), div2(z0))
cond2(false, z0) → cond1(neq(z0, 0), p(z0))
neq(0, 0) → false
neq(0, s(z0)) → true
neq(s(z0), 0) → true
neq(s(z0), s(y)) → neq(z0, y)
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
COND1(true, z0) → c(COND2(even(z0), z0), EVEN(z0))
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
S tuples:
COND1(true, z0) → c(COND2(even(z0), z0), EVEN(z0))
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
K tuples:none
Defined Rule Symbols:
cond1, cond2, neq, even, div2, p
Defined Pair Symbols:
COND1, EVEN, DIV2, COND2
Compound Symbols:
c, c9, c12, c1, c2
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
cond1(true, z0) → cond2(even(z0), z0)
cond2(true, z0) → cond1(neq(z0, 0), div2(z0))
cond2(false, z0) → cond1(neq(z0, 0), p(z0))
neq(0, s(z0)) → true
neq(s(z0), s(y)) → neq(z0, y)
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
COND1(true, z0) → c(COND2(even(z0), z0), EVEN(z0))
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
S tuples:
COND1(true, z0) → c(COND2(even(z0), z0), EVEN(z0))
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
K tuples:none
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
COND1, EVEN, DIV2, COND2
Compound Symbols:
c, c9, c12, c1, c2
(9) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
COND1(
true,
z0) →
c(
COND2(
even(
z0),
z0),
EVEN(
z0)) by
COND1(true, 0) → c(COND2(true, 0), EVEN(0))
COND1(true, s(0)) → c(COND2(false, s(0)), EVEN(s(0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, 0) → c(COND2(true, 0), EVEN(0))
COND1(true, s(0)) → c(COND2(false, s(0)), EVEN(s(0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, 0) → c(COND2(true, 0), EVEN(0))
COND1(true, s(0)) → c(COND2(false, s(0)), EVEN(s(0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
K tuples:none
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND2, COND1
Compound Symbols:
c9, c12, c1, c2, c
(11) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing tuple parts
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
K tuples:none
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND2, COND1
Compound Symbols:
c9, c12, c1, c2, c, c
(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
COND1(true, 0) → c(COND2(true, 0))
We considered the (Usable) Rules:
neq(s(z0), 0) → true
neq(0, 0) → false
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(COND1(x1, x2)) = x1
POL(COND2(x1, x2)) = x2
POL(DIV2(x1)) = 0
POL(EVEN(x1)) = 0
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = 0
POL(even(x1)) = 0
POL(false) = 0
POL(neq(x1, x2)) = x1
POL(p(x1)) = 0
POL(s(x1)) = [1]
POL(true) = [1]
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(true, z0) → c1(COND1(neq(z0, 0), div2(z0)), DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND2, COND1
Compound Symbols:
c9, c12, c1, c2, c, c
(15) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
COND2(
true,
z0) →
c1(
COND1(
neq(
z0,
0),
div2(
z0)),
DIV2(
z0)) by
COND2(true, 0) → c1(COND1(neq(0, 0), 0), DIV2(0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0), DIV2(s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, 0) → c1(COND1(false, div2(0)), DIV2(0))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
(16) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0), DIV2(0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0), DIV2(s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, 0) → c1(COND1(false, div2(0)), DIV2(0))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0), DIV2(0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0), DIV2(s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, 0) → c1(COND1(false, div2(0)), DIV2(0))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND2, COND1
Compound Symbols:
c9, c12, c2, c, c, c1
(17) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
COND2(true, 0) → c1(COND1(false, div2(0)), DIV2(0))
(18) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0), DIV2(0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0), DIV2(s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0), DIV2(0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0), DIV2(s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND2, COND1
Compound Symbols:
c9, c12, c2, c, c, c1
(19) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing tuple parts
(20) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND2, COND1
Compound Symbols:
c9, c12, c2, c, c, c1, c1
(21) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)
The following tuples could be moved from S to K by knowledge propagation:
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, 0) → c(COND2(true, 0))
(22) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND2, COND1
Compound Symbols:
c9, c12, c2, c, c, c1, c1
(23) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
We considered the (Usable) Rules:
div2(0) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
div2(s(0)) → 0
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [2]
POL(COND1(x1, x2)) = x2
POL(COND2(x1, x2)) = x2
POL(DIV2(x1)) = 0
POL(EVEN(x1)) = 0
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = x1
POL(even(x1)) = [2] + x1
POL(false) = 0
POL(neq(x1, x2)) = 0
POL(p(x1)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0
(24) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND2(false, z0) → c2(COND1(neq(z0, 0), p(z0)))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND2, COND1
Compound Symbols:
c9, c12, c2, c, c, c1, c1
(25) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
COND2(
false,
z0) →
c2(
COND1(
neq(
z0,
0),
p(
z0))) by
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, 0) → c2(COND1(false, p(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
(26) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, 0) → c2(COND1(false, p(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, 0) → c2(COND1(false, p(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c2
(27) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
COND2(false, 0) → c2(COND1(false, p(0)))
(28) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c2
(29) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)
The following tuples could be moved from S to K by knowledge propagation:
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND1(true, 0) → c(COND2(true, 0))
(30) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(0, 0) → false
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(0) → 0
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c2
(31) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
p(0) → 0
(32) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c2
(33) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
We considered the (Usable) Rules:
div2(0) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
div2(s(0)) → 0
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [1]
POL(COND1(x1, x2)) = x2
POL(COND2(x1, x2)) = x2
POL(DIV2(x1)) = 0
POL(EVEN(x1)) = 0
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = x1
POL(even(x1)) = 0
POL(false) = 0
POL(neq(x1, x2)) = 0
POL(p(x1)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0
(34) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(z0))) → c(COND2(even(z0), s(s(z0))), EVEN(s(s(z0))))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c2
(35) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
COND1(
true,
s(
s(
z0))) →
c(
COND2(
even(
z0),
s(
s(
z0))),
EVEN(
s(
s(
z0)))) by
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
(36) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c1, c2, c
(37) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
We considered the (Usable) Rules:none
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [1]
POL(COND1(x1, x2)) = [1]
POL(COND2(x1, x2)) = [1]
POL(DIV2(x1)) = 0
POL(EVEN(x1)) = 0
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = [1] + x1
POL(even(x1)) = [1]
POL(false) = 0
POL(neq(x1, x2)) = [1] + x1 + x2
POL(p(x1)) = 0
POL(s(x1)) = [1]
POL(true) = 0
(38) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c1, c2, c
(39) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
COND2(
true,
s(
s(
z0))) →
c1(
COND1(
neq(
s(
s(
z0)),
0),
s(
div2(
z0))),
DIV2(
s(
s(
z0)))) by
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(0)))) → c1(COND1(neq(s(s(s(0))), 0), s(0)), DIV2(s(s(s(0)))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
(40) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(0)))) → c1(COND1(neq(s(s(s(0))), 0), s(0)), DIV2(s(s(s(0)))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c1, c2, c
(41) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)
Split RHS of tuples not part of any SCC
(42) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(true, s(s(s(0)))) → c3(DIV2(s(s(s(0)))))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, s(s(z0))) → c1(COND1(neq(s(s(z0)), 0), s(div2(z0))), DIV2(s(s(z0))))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c1, c2, c, c3
(43) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
COND2(true, s(s(s(0)))) → c3(DIV2(s(s(s(0)))))
(44) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(z0)) → c1(COND1(true, div2(s(z0))), DIV2(s(z0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c1, c2, c, c3
(45) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
COND2(
true,
s(
z0)) →
c1(
COND1(
true,
div2(
s(
z0))),
DIV2(
s(
z0))) by
COND2(true, s(0)) → c1(COND1(true, 0), DIV2(s(0)))
COND2(true, s(s(z0))) → c1(COND1(true, s(div2(z0))), DIV2(s(s(z0))))
(46) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(true, s(0)) → c1(COND1(true, 0), DIV2(s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND2(true, s(0)) → c1(COND1(true, 0), DIV2(s(0)))
COND2(true, s(s(z0))) → c1(COND1(true, s(div2(z0))), DIV2(s(s(z0))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c2, c, c1, c3
(47) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
COND2(true, s(0)) → c1(COND1(true, 0), DIV2(s(0)))
(48) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND2(true, s(s(z0))) → c1(COND1(true, s(div2(z0))), DIV2(s(s(z0))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c2, c, c1, c3
(49) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
We considered the (Usable) Rules:
div2(0) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
div2(s(0)) → 0
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(COND1(x1, x2)) = x2
POL(COND2(x1, x2)) = x2
POL(DIV2(x1)) = 0
POL(EVEN(x1)) = 0
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = x1
POL(even(x1)) = 0
POL(false) = 0
POL(neq(x1, x2)) = [1]
POL(p(x1)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0
(50) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c2, c, c1, c3
(51) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
DIV2(s(s(z0))) → c12(DIV2(z0))
We considered the (Usable) Rules:
div2(0) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
div2(s(0)) → 0
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [2]
POL(COND1(x1, x2)) = [2]x2 + [2]x22
POL(COND2(x1, x2)) = [2]x2 + [2]x22
POL(DIV2(x1)) = [2] + [2]x1
POL(EVEN(x1)) = 0
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = [1] + x1
POL(even(x1)) = 0
POL(false) = 0
POL(neq(x1, x2)) = x2 + [2]x22
POL(p(x1)) = x1
POL(s(x1)) = [2] + x1
POL(true) = 0
(52) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c2, c, c1, c3
(53) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
COND2(
true,
0) →
c1(
COND1(
neq(
0,
0),
0)) by
COND2(true, 0) → c1(COND1(false, 0))
(54) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, 0) → c(COND2(true, 0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(true, 0) → c1(COND1(false, 0))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND1(true, 0) → c(COND2(true, 0))
COND2(true, 0) → c1(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c1, c2, c, c1, c3
(55) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 4 trailing nodes:
COND1(true, 0) → c(COND2(true, 0))
COND2(false, 0) → c2(COND1(neq(0, 0), 0))
COND2(true, s(0)) → c1(COND1(neq(s(0), 0), 0))
COND2(true, 0) → c1(COND1(false, 0))
(56) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
neq(0, 0) → false
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
p(s(z0)) → z0
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
Defined Rule Symbols:
even, neq, div2, p
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c2, c, c1, c1, c3
(57) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
neq(0, 0) → false
(58) Obligation:
Complexity Dependency Tuples Problem
Rules:
neq(s(z0), 0) → true
p(s(z0)) → z0
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
Defined Rule Symbols:
neq, p, even, div2
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c2, c, c1, c1, c3
(59) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
We considered the (Usable) Rules:
div2(0) → 0
div2(s(s(z0))) → s(div2(z0))
even(0) → true
p(s(z0)) → z0
even(s(0)) → false
div2(s(0)) → 0
even(s(s(z0))) → even(z0)
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(COND1(x1, x2)) = [1] + x2
POL(COND2(x1, x2)) = x1 + x2
POL(DIV2(x1)) = 0
POL(EVEN(x1)) = 0
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = x1
POL(even(x1)) = [1]
POL(false) = [1]
POL(neq(x1, x2)) = 0
POL(p(x1)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0
(60) Obligation:
Complexity Dependency Tuples Problem
Rules:
neq(s(z0), 0) → true
p(s(z0)) → z0
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
Defined Rule Symbols:
neq, p, even, div2
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c2, c, c1, c1, c3
(61) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
COND2(
false,
s(
z0)) →
c2(
COND1(
neq(
s(
z0),
0),
z0)) by
COND2(false, s(z0)) → c2(COND1(true, z0))
(62) Obligation:
Complexity Dependency Tuples Problem
Rules:
neq(s(z0), 0) → true
p(s(z0)) → z0
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(false, s(z0)) → c2(COND1(true, z0))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND2(false, s(z0)) → c2(COND1(true, p(s(z0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
Defined Rule Symbols:
neq, p, even, div2
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c2, c, c1, c1, c3
(63) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
COND2(
false,
s(
z0)) →
c2(
COND1(
true,
p(
s(
z0)))) by
COND2(false, s(z0)) → c2(COND1(true, z0))
(64) Obligation:
Complexity Dependency Tuples Problem
Rules:
neq(s(z0), 0) → true
p(s(z0)) → z0
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(false, s(z0)) → c2(COND1(true, z0))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND2(false, s(z0)) → c2(COND1(true, z0))
K tuples:
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
Defined Rule Symbols:
neq, p, even, div2
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c3, c2
(65) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
p(s(z0)) → z0
(66) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(false, s(z0)) → c2(COND1(true, z0))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND2(false, s(z0)) → c2(COND1(true, z0))
K tuples:
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
Defined Rule Symbols:
even, neq, div2
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c3, c2
(67) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
COND2(false, s(z0)) → c2(COND1(true, z0))
We considered the (Usable) Rules:
div2(0) → 0
div2(s(s(z0))) → s(div2(z0))
div2(s(0)) → 0
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(false, s(z0)) → c2(COND1(true, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [1]
POL(COND1(x1, x2)) = x2
POL(COND2(x1, x2)) = x2
POL(DIV2(x1)) = [1]
POL(EVEN(x1)) = 0
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = x1
POL(even(x1)) = 0
POL(false) = 0
POL(neq(x1, x2)) = 0
POL(s(x1)) = [1] + x1
POL(true) = 0
(68) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(false, s(z0)) → c2(COND1(true, z0))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
K tuples:
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND2(false, s(z0)) → c2(COND1(true, z0))
Defined Rule Symbols:
even, neq, div2
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c3, c2
(69) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
We considered the (Usable) Rules:
div2(0) → 0
neq(s(z0), 0) → true
div2(s(s(z0))) → s(div2(z0))
div2(s(0)) → 0
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(false, s(z0)) → c2(COND1(true, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(COND1(x1, x2)) = x1 + x2
POL(COND2(x1, x2)) = x2
POL(DIV2(x1)) = 0
POL(EVEN(x1)) = 0
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = x1
POL(even(x1)) = 0
POL(false) = 0
POL(neq(x1, x2)) = [1]
POL(s(x1)) = [1] + x1
POL(true) = [1]
(70) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(false, s(z0)) → c2(COND1(true, z0))
S tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
K tuples:
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND2(false, s(z0)) → c2(COND1(true, z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
Defined Rule Symbols:
even, neq, div2
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c3, c2
(71) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
EVEN(s(s(z0))) → c9(EVEN(z0))
We considered the (Usable) Rules:
div2(0) → 0
neq(s(z0), 0) → true
div2(s(s(z0))) → s(div2(z0))
div2(s(0)) → 0
And the Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(false, s(z0)) → c2(COND1(true, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [1]
POL(COND1(x1, x2)) = [2]x22 + [2]x1·x2
POL(COND2(x1, x2)) = [2] + [2]x22
POL(DIV2(x1)) = 0
POL(EVEN(x1)) = [2]x1
POL(c(x1)) = x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(c1(x1, x2)) = x1 + x2
POL(c12(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(c9(x1)) = x1
POL(div2(x1)) = x1
POL(even(x1)) = 0
POL(false) = 0
POL(neq(x1, x2)) = [2]x22
POL(s(x1)) = [1] + x1
POL(true) = [2]
(72) Obligation:
Complexity Dependency Tuples Problem
Rules:
even(0) → true
even(s(0)) → false
even(s(s(z0))) → even(z0)
neq(s(z0), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(z0))) → s(div2(z0))
Tuples:
EVEN(s(s(z0))) → c9(EVEN(z0))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(0))) → c1(COND1(neq(s(s(0)), 0), s(0)), DIV2(s(s(0))))
COND2(true, s(s(s(s(z0))))) → c1(COND1(neq(s(s(s(s(z0)))), 0), s(s(div2(z0)))), DIV2(s(s(s(s(z0))))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
COND2(true, s(s(x0))) → c1(DIV2(s(s(x0))))
COND2(true, s(s(s(0)))) → c3(COND1(neq(s(s(s(0))), 0), s(0)))
COND2(false, s(z0)) → c2(COND1(true, z0))
S tuples:none
K tuples:
COND2(false, s(z0)) → c2(COND1(neq(s(z0), 0), z0))
COND1(true, s(s(x0))) → c(EVEN(s(s(x0))))
COND2(true, s(s(x0))) → c1(COND1(true, s(div2(x0))), DIV2(s(s(x0))))
DIV2(s(s(z0))) → c12(DIV2(z0))
COND1(true, s(s(0))) → c(COND2(true, s(s(0))), EVEN(s(s(0))))
COND2(false, s(z0)) → c2(COND1(true, z0))
COND1(true, s(0)) → c(COND2(false, s(0)))
COND1(true, s(s(s(0)))) → c(COND2(false, s(s(s(0)))), EVEN(s(s(s(0)))))
COND1(true, s(s(s(s(z0))))) → c(COND2(even(z0), s(s(s(s(z0))))), EVEN(s(s(s(s(z0))))))
EVEN(s(s(z0))) → c9(EVEN(z0))
Defined Rule Symbols:
even, neq, div2
Defined Pair Symbols:
EVEN, DIV2, COND1, COND2
Compound Symbols:
c9, c12, c, c, c1, c1, c3, c2
(73) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(74) BOUNDS(1, 1)