Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
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
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
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
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is
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
The TRS R 2 is
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))
The signature Sigma is {cond2, cond1}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
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
The set Q consists of the following terms:
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
DIV2(s(s(x))) → DIV2(x)
EVEN(s(s(x))) → EVEN(x)
COND1(true, x) → COND2(even(x), x)
NEQ(s(x), s(y)) → NEQ(x, y)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(true, x) → COND1(neq(x, 0), div2(x))
COND2(false, x) → NEQ(x, 0)
COND1(true, x) → EVEN(x)
COND2(true, x) → DIV2(x)
COND2(false, x) → P(x)
COND2(true, x) → NEQ(x, 0)
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
The set Q consists of the following terms:
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
DIV2(s(s(x))) → DIV2(x)
EVEN(s(s(x))) → EVEN(x)
COND1(true, x) → COND2(even(x), x)
NEQ(s(x), s(y)) → NEQ(x, y)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(true, x) → COND1(neq(x, 0), div2(x))
COND2(false, x) → NEQ(x, 0)
COND1(true, x) → EVEN(x)
COND2(true, x) → DIV2(x)
COND2(false, x) → P(x)
COND2(true, x) → NEQ(x, 0)
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
The set Q consists of the following terms:
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 6 less nodes.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
DIV2(s(s(x))) → DIV2(x)
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
The set Q consists of the following terms:
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
DIV2(s(s(x))) → DIV2(x)
R is empty.
The set Q consists of the following terms:
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
DIV2(s(s(x))) → DIV2(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- DIV2(s(s(x))) → DIV2(x)
The graph contains the following edges 1 > 1
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
EVEN(s(s(x))) → EVEN(x)
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
The set Q consists of the following terms:
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
EVEN(s(s(x))) → EVEN(x)
R is empty.
The set Q consists of the following terms:
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
EVEN(s(s(x))) → EVEN(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- EVEN(s(s(x))) → EVEN(x)
The graph contains the following edges 1 > 1
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(true, x) → COND1(neq(x, 0), div2(x))
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
The set Q consists of the following terms:
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(true, x) → COND1(neq(x, 0), div2(x))
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(true, x) → COND1(neq(x, 0), div2(x))
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND2(false, x) → COND1(neq(x, 0), p(x)) at position [0] we obtained the following new rules:
COND2(false, 0) → COND1(false, p(0))
COND2(false, s(x0)) → COND1(true, p(s(x0)))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(false, 0) → COND1(false, p(0))
COND2(false, s(x0)) → COND1(true, p(s(x0)))
COND2(true, x) → COND1(neq(x, 0), div2(x))
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(false, s(x0)) → COND1(true, p(s(x0)))
COND2(true, x) → COND1(neq(x, 0), div2(x))
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(false, s(x0)) → COND1(true, p(s(x0)))
COND2(true, x) → COND1(neq(x, 0), div2(x))
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
p(s(x)) → x
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND2(false, s(x0)) → COND1(true, p(s(x0))) at position [1] we obtained the following new rules:
COND2(false, s(x0)) → COND1(true, x0)
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(true, x) → COND1(neq(x, 0), div2(x))
COND2(false, s(x0)) → COND1(true, x0)
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
p(s(x)) → x
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(true, x) → COND1(neq(x, 0), div2(x))
COND2(false, s(x0)) → COND1(true, x0)
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
p(0)
p(s(x0))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(true, x) → COND1(neq(x, 0), div2(x))
COND2(false, s(x0)) → COND1(true, x0)
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND2(true, x) → COND1(neq(x, 0), div2(x)) at position [0] we obtained the following new rules:
COND2(true, s(x0)) → COND1(true, div2(s(x0)))
COND2(true, 0) → COND1(false, div2(0))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(true, 0) → COND1(false, div2(0))
COND2(true, s(x0)) → COND1(true, div2(s(x0)))
COND2(false, s(x0)) → COND1(true, x0)
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(true, s(x0)) → COND1(true, div2(s(x0)))
COND2(false, s(x0)) → COND1(true, x0)
The TRS R consists of the following rules:
neq(0, 0) → false
neq(s(x), 0) → true
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(true, s(x0)) → COND1(true, div2(s(x0)))
COND2(false, s(x0)) → COND1(true, x0)
The TRS R consists of the following rules:
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
div2(0) → 0
The set Q consists of the following terms:
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(true, s(x0)) → COND1(true, div2(s(x0)))
COND2(false, s(x0)) → COND1(true, x0)
The TRS R consists of the following rules:
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
div2(0) → 0
The set Q consists of the following terms:
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
COND2(false, s(x0)) → COND1(true, x0)
The remaining pairs can at least be oriented weakly.
COND1(true, x) → COND2(even(x), x)
COND2(true, s(x0)) → COND1(true, div2(s(x0)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 1
POL(COND1(x1, x2)) = x1 + x2
POL(COND2(x1, x2)) = 1 + x2
POL(div2(x1)) = x1
POL(even(x1)) = 0
POL(false) = 0
POL(s(x1)) = 1 + x1
POL(true) = 1
The following usable rules [17] were oriented:
div2(0) → 0
div2(s(s(x))) → s(div2(x))
div2(s(0)) → 0
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
COND2(true, s(x0)) → COND1(true, div2(s(x0)))
The TRS R consists of the following rules:
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
div2(0) → 0
The set Q consists of the following terms:
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
COND2(true, s(x0)) → COND1(true, div2(s(x0)))
The remaining pairs can at least be oriented weakly.
COND1(true, x) → COND2(even(x), x)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
Tuple symbols:
| M( COND1(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
| M( COND2(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
div2(0) → 0
div2(s(s(x))) → s(div2(x))
div2(s(0)) → 0
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
COND1(true, x) → COND2(even(x), x)
The TRS R consists of the following rules:
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
div2(0) → 0
The set Q consists of the following terms:
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.