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:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
F(true, x, y) → GT(x, y)
GT(s(u), s(v)) → GT(u, v)
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
F(true, x, y) → GT(x, y)
GT(s(u), s(v)) → GT(u, v)
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
GT(s(u), s(v)) → GT(u, v)
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)
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 [13] we can delete all non-usable rules [14] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
GT(s(u), s(v)) → GT(u, v)
R is empty.
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)
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.
gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
GT(s(u), s(v)) → GT(u, v)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] 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:
- GT(s(u), s(v)) → GT(u, v)
The graph contains the following edges 1 > 1, 2 > 2
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)
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 [13] we can delete all non-usable rules [14] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)
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.
f(true, x0, x1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule F(true, x, y) → F(gt(x, y), s(x), s(s(y))) at position [0] we obtained the following new rules:
F(true, 0, x0) → F(false, s(0), s(s(x0)))
F(true, s(x0), s(x1)) → F(gt(x0, x1), s(s(x0)), s(s(s(x1))))
F(true, s(x0), 0) → F(true, s(s(x0)), s(s(0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
F(true, 0, x0) → F(false, s(0), s(s(x0)))
F(true, s(x0), 0) → F(true, s(s(x0)), s(s(0)))
F(true, s(x0), s(x1)) → F(gt(x0, x1), s(s(x0)), s(s(s(x1))))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
F(true, s(x0), s(x1)) → F(gt(x0, x1), s(s(x0)), s(s(s(x1))))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule F(true, s(x0), s(x1)) → F(gt(x0, x1), s(s(x0)), s(s(s(x1)))) we obtained the following new rules:
F(true, s(s(z0)), s(s(s(z1)))) → F(gt(s(z0), s(s(z1))), s(s(s(z0))), s(s(s(s(s(z1))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
F(true, s(s(z0)), s(s(s(z1)))) → F(gt(s(z0), s(s(z1))), s(s(s(z0))), s(s(s(s(s(z1))))))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule F(true, s(s(z0)), s(s(s(z1)))) → F(gt(s(z0), s(s(z1))), s(s(s(z0))), s(s(s(s(s(z1)))))) at position [0] we obtained the following new rules:
F(true, s(s(z0)), s(s(s(z1)))) → F(gt(z0, s(z1)), s(s(s(z0))), s(s(s(s(s(z1))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Rewriting
↳ QDP
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
F(true, s(s(z0)), s(s(s(z1)))) → F(gt(z0, s(z1)), s(s(s(z0))), s(s(s(s(s(z1))))))
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus with the following steps:
Note that final constraints are written in bold face.
For Pair F(true, x, y) → F(gt(x, y), s(x), s(s(y))) the following chains were created:
- We consider the chain F(true, x, y) → F(gt(x, y), s(x), s(s(y))), F(true, x, y) → F(gt(x, y), s(x), s(s(y))) which results in the following constraint:
- (1) (F(gt(x0, x1), s(x0), s(s(x1)))=F(true, x2, x3) ⇒ F(true, x0, x1)≥F(gt(x0, x1), s(x0), s(s(x1))))
We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
- (2) (gt(x0, x1)=true ⇒ F(true, x0, x1)≥F(gt(x0, x1), s(x0), s(s(x1))))
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x0, x1)=true which results in the following new constraints:
- (3) (true=true ⇒ F(true, s(x5), 0)≥F(gt(s(x5), 0), s(s(x5)), s(s(0))))
- (4) (gt(x6, x7)=true∧(gt(x6, x7)=true ⇒ F(true, x6, x7)≥F(gt(x6, x7), s(x6), s(s(x7)))) ⇒ F(true, s(x6), s(x7))≥F(gt(s(x6), s(x7)), s(s(x6)), s(s(s(x7)))))
We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
- (5) (F(true, s(x5), 0)≥F(gt(s(x5), 0), s(s(x5)), s(s(0))))
We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gt(x6, x7)=true ⇒ F(true, x6, x7)≥F(gt(x6, x7), s(x6), s(s(x7)))) with σ = [ ] which results in the following new constraint:
- (6) (F(true, x6, x7)≥F(gt(x6, x7), s(x6), s(s(x7))) ⇒ F(true, s(x6), s(x7))≥F(gt(s(x6), s(x7)), s(s(x6)), s(s(s(x7)))))
To summarize, we get the following constraints P≥ for the following pairs.
- F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
- (F(true, s(x5), 0)≥F(gt(s(x5), 0), s(s(x5)), s(s(0))))
- (F(true, x6, x7)≥F(gt(x6, x7), s(x6), s(s(x7))) ⇒ F(true, s(x6), s(x7))≥F(gt(s(x6), s(x7)), s(s(x6)), s(s(s(x7)))))
The constraints for P> respective Pbound are constructed from P≥ where we just replace every occurence of "t ≥ s" in P≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved.
Polynomial interpretation [21]:
POL(0) = 0
POL(F(x1, x2, x3)) = -1 - x1 + x2 - x3
POL(c) = -1
POL(false) = 1
POL(gt(x1, x2)) = 0
POL(s(x1)) = 2 + x1
POL(true) = 0
The following pairs are in P>:
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
The following pairs are in Pbound:
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
The following rules are usable:
true → gt(s(u), 0)
false → gt(0, v)
gt(u, v) → gt(s(u), s(v))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ NonInfProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.