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(f(a, x), y) → f(a, f(f(x, h(y)), a))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x), y) → F(x, h(y))
F(f(a, x), y) → F(f(x, h(y)), a)
F(f(a, x), y) → F(a, f(f(x, h(y)), a))
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x), y) → F(x, h(y))
F(f(a, x), y) → F(f(x, h(y)), a)
F(f(a, x), y) → F(a, f(f(x, h(y)), a))
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x), y) → F(x, h(y))
F(f(a, x), y) → F(f(x, h(y)), a)
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(f(a, x), y) → F(x, h(y)) we obtained the following new rules:
F(f(a, x0), a) → F(x0, h(a))
F(f(a, x0), h(z1)) → F(x0, h(h(z1)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x0), a) → F(x0, h(a))
F(f(a, x), y) → F(f(x, h(y)), a)
F(f(a, x0), h(z1)) → F(x0, h(h(z1)))
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(f(a, x), y) → F(f(x, h(y)), a) we obtained the following new rules:
F(f(a, x0), h(h(z1))) → F(f(x0, h(h(h(z1)))), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), a) → F(f(x0, h(a)), a)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x0), a) → F(x0, h(a))
F(f(a, x0), h(h(z1))) → F(f(x0, h(h(h(z1)))), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), a) → F(f(x0, h(a)), a)
F(f(a, x0), h(z1)) → F(x0, h(h(z1)))
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(f(a, x0), h(z1)) → F(x0, h(h(z1))) we obtained the following new rules:
F(f(a, x0), h(a)) → F(x0, h(h(a)))
F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x0), h(a)) → F(x0, h(h(a)))
F(f(a, x0), a) → F(x0, h(a))
F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1))))
F(f(a, x0), h(h(z1))) → F(f(x0, h(h(h(z1)))), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), a) → F(f(x0, h(a)), a)
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(f(a, x0), h(h(z1))) → F(f(x0, h(h(h(z1)))), a) we obtained the following new rules:
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x0), h(a)) → F(x0, h(h(a)))
F(f(a, x0), a) → F(x0, h(a))
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)
F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1))))
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), a) → F(f(x0, h(a)), a)
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(f(a, x0), a) → F(x0, h(a)) we obtained the following new rules:
F(f(a, f(a, y_0)), a) → F(f(a, y_0), h(a))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x0), h(a)) → F(x0, h(h(a)))
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)
F(f(a, f(a, y_0)), a) → F(f(a, y_0), h(a))
F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1))))
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), a) → F(f(x0, h(a)), a)
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(f(a, x0), h(a)) → F(x0, h(h(a))) we obtained the following new rules:
F(f(a, f(a, y_0)), h(a)) → F(f(a, y_0), h(h(a)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)
F(f(a, f(a, y_0)), a) → F(f(a, y_0), h(a))
F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1))))
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, f(a, y_0)), h(a)) → F(f(a, y_0), h(h(a)))
F(f(a, x0), a) → F(f(x0, h(a)), a)
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1)))) we obtained the following new rules:
F(f(a, f(a, y_0)), h(h(x1))) → F(f(a, y_0), h(h(h(x1))))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ MNOCProof
↳ SemLabProof
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)
F(f(a, f(a, y_0)), a) → F(f(a, y_0), h(a))
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, f(a, y_0)), h(h(x1))) → F(f(a, y_0), h(h(h(x1))))
F(f(a, f(a, y_0)), h(a)) → F(f(a, y_0), h(h(a)))
F(f(a, x0), a) → F(f(x0, h(a)), a)
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
The set Q consists of the following terms:
f(f(a, x0), x1)
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ MNOCProof
↳ QDP
↳ SemLabProof
Q DP problem:
The TRS P consists of the following rules:
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)
F(f(a, f(a, y_0)), a) → F(f(a, y_0), h(a))
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, f(a, y_0)), h(a)) → F(f(a, y_0), h(h(a)))
F(f(a, f(a, y_0)), h(h(x1))) → F(f(a, y_0), h(h(h(x1))))
F(f(a, x0), a) → F(f(x0, h(a)), a)
The TRS R consists of the following rules:
f(f(a, x), y) → f(a, f(f(x, h(y)), a))
Q is empty.
We have to consider all (P,Q,R)-chains.
We found the following model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.a: 1
f: 0
h: 0
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
F.0-0(f.1-1(a., x0), h.0(h.1(a.))) → F.0-1(f.1-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.0(x1))))
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-1(a., x0), h.1(a.)) → F.0-1(f.1-0(x0, h.0(h.1(a.))), a.)
F.0-1(f.1-0(a., f.1-1(a., y_0)), a.) → F.0-0(f.1-1(a., y_0), h.1(a.))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-1(f.1-1(a., x0), a.) → F.0-1(f.1-0(x0, h.1(a.)), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.1(a.)) → F.0-0(f.1-1(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
The TRS R consists of the following rules:
f.0-0(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.0(y)), a.))
f.0-1(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.1(y)), a.))
f.0-0(f.1-1(a., x), y) → f.1-0(a., f.0-1(f.1-0(x, h.0(y)), a.))
f.0-1(f.1-1(a., x), y) → f.1-0(a., f.0-1(f.1-0(x, h.1(y)), a.))
The set Q consists of the following terms:
f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ MNOCProof
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F.0-0(f.1-1(a., x0), h.0(h.1(a.))) → F.0-1(f.1-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.0(x1))))
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-1(a., x0), h.1(a.)) → F.0-1(f.1-0(x0, h.0(h.1(a.))), a.)
F.0-1(f.1-0(a., f.1-1(a., y_0)), a.) → F.0-0(f.1-1(a., y_0), h.1(a.))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-1(f.1-1(a., x0), a.) → F.0-1(f.1-0(x0, h.1(a.)), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.1(a.)) → F.0-0(f.1-1(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
The TRS R consists of the following rules:
f.0-0(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.0(y)), a.))
f.0-1(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.1(y)), a.))
f.0-0(f.1-1(a., x), y) → f.1-0(a., f.0-1(f.1-0(x, h.0(y)), a.))
f.0-1(f.1-1(a., x), y) → f.1-0(a., f.0-1(f.1-0(x, h.1(y)), a.))
The set Q consists of the following terms:
f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ MNOCProof
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
F.0-0(f.1-1(a., x0), h.0(h.1(a.))) → F.0-1(f.1-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.0(x1))))
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-1(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.1(a.)) → F.0-0(f.1-1(a., y_0), h.0(h.1(a.)))
F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
The TRS R consists of the following rules:
f.0-0(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.0(y)), a.))
f.0-1(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.1(y)), a.))
f.0-0(f.1-1(a., x), y) → f.1-0(a., f.0-1(f.1-0(x, h.0(y)), a.))
f.0-1(f.1-1(a., x), y) → f.1-0(a., f.0-1(f.1-0(x, h.1(y)), a.))
The set Q consists of the following terms:
f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
F.0-0(f.1-1(a., x0), h.0(h.1(a.))) → F.0-1(f.1-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
Strictly oriented rules of the TRS R:
f.0-0(f.1-1(a., x), y) → f.1-0(a., f.0-1(f.1-0(x, h.0(y)), a.))
f.0-1(f.1-1(a., x), y) → f.1-0(a., f.0-1(f.1-0(x, h.1(y)), a.))
Used ordering: POLO with Polynomial interpretation [25]:
POL(F.0-0(x1, x2)) = x1 + x2
POL(F.0-1(x1, x2)) = x1 + x2
POL(a.) = 0
POL(f.0-0(x1, x2)) = x1 + x2
POL(f.0-1(x1, x2)) = x1 + x2
POL(f.1-0(x1, x2)) = x1 + x2
POL(f.1-1(x1, x2)) = 1 + x1 + x2
POL(h.0(x1)) = x1
POL(h.1(x1)) = x1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ MNOCProof
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.1(a.)) → F.0-0(f.1-1(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.0(x1))))
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
The TRS R consists of the following rules:
f.0-0(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.0(y)), a.))
f.0-1(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.1(y)), a.))
The set Q consists of the following terms:
f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ MNOCProof
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
The TRS R consists of the following rules:
f.0-0(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.0(y)), a.))
f.0-1(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.1(y)), a.))
The set Q consists of the following terms:
f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
Used ordering: POLO with Polynomial interpretation [25]:
POL(F.0-0(x1, x2)) = 1 + x1 + x2
POL(F.0-1(x1, x2)) = 1 + x1 + x2
POL(a.) = 0
POL(f.0-0(x1, x2)) = x1 + x2
POL(f.0-1(x1, x2)) = x1 + x2
POL(f.1-0(x1, x2)) = 1 + x1 + x2
POL(h.0(x1)) = x1
POL(h.1(x1)) = x1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ MNOCProof
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f.0-0(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.0(y)), a.))
f.0-1(f.1-0(a., x), y) → f.1-0(a., f.0-1(f.0-0(x, h.1(y)), a.))
The set Q consists of the following terms:
f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.