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(x, f(y, a)) → f(f(a, f(h(x), y)), a)
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), 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(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
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(x, f(y, a)) → F(h(x), y)
F(x, f(y, a)) → F(a, f(h(x), y))
F(x, f(y, a)) → F(f(a, f(h(x), y)), a)
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
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(x, f(y, a)) → F(h(x), y)
F(x, f(y, a)) → F(a, f(h(x), y))
F(x, f(y, a)) → F(f(a, f(h(x), y)), a)
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
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(x, f(y, a)) → F(h(x), y)
F(x, f(y, a)) → F(a, f(h(x), y))
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, f(y, a)) → F(h(x), y) we obtained the following new rules:
F(a, f(x1, a)) → F(h(a), x1)
F(h(z0), f(x1, a)) → F(h(h(z0)), x1)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(a, f(x1, a)) → F(h(a), x1)
F(h(z0), f(x1, a)) → F(h(h(z0)), x1)
F(x, f(y, a)) → F(a, f(h(x), y))
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, f(y, a)) → F(a, f(h(x), y)) we obtained the following new rules:
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(h(z0)), f(x1, a)) → F(a, f(h(h(h(z0))), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))
↳ 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(a, f(x1, a)) → F(h(a), x1)
F(h(z0), f(x1, a)) → F(h(h(z0)), x1)
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))
F(h(h(z0)), f(x1, a)) → F(a, f(h(h(h(z0))), x1))
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(h(z0), f(x1, a)) → F(h(h(z0)), x1) we obtained the following new rules:
F(h(a), f(x1, a)) → F(h(h(a)), x1)
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)
↳ 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(h(a), f(x1, a)) → F(h(h(a)), x1)
F(a, f(x1, a)) → F(h(a), x1)
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(h(z0)), f(x1, a)) → F(a, f(h(h(h(z0))), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(h(h(z0)), f(x1, a)) → F(a, f(h(h(h(z0))), x1)) we obtained the following new rules:
F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
↳ 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(h(a), f(x1, a)) → F(h(h(a)), x1)
F(a, f(x1, a)) → F(h(a), x1)
F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(a, f(x1, a)) → F(h(a), x1) we obtained the following new rules:
F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, 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(h(a), f(x1, a)) → F(h(h(a)), x1)
F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, a))
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(h(a), f(x1, a)) → F(h(h(a)), x1) we obtained the following new rules:
F(h(a), f(f(y_1, a), a)) → F(h(h(a)), f(y_1, 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(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)
F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, a))
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(a), f(f(y_1, a), a)) → F(h(h(a)), f(y_1, a))
F(a, f(x1, a)) → F(a, f(h(a), x1))
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1) we obtained the following new rules:
F(h(h(x0)), f(f(y_1, a), a)) → F(h(h(h(x0))), f(y_1, a))
↳ 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(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, a))
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(h(x0)), f(f(y_1, a), a)) → F(h(h(h(x0))), f(y_1, a))
F(h(a), f(f(y_1, a), a)) → F(h(h(a)), f(y_1, a))
F(a, f(x1, a)) → F(a, f(h(a), x1))
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), y)), a)
The set Q consists of the following terms:
f(x0, f(x1, a))
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(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, a))
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(a), f(f(y_1, a), a)) → F(h(h(a)), f(y_1, a))
F(h(h(x0)), f(f(y_1, a), a)) → F(h(h(h(x0))), f(y_1, a))
F(a, f(x1, a)) → F(a, f(h(a), x1))
The TRS R consists of the following rules:
f(x, f(y, a)) → f(f(a, f(h(x), 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(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.0-0(h.1(a.), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
F.0-0(h.1(a.), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.1(a.)), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
F.1-0(a., f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.1(a.), x1))
F.1-0(a., f.0-1(f.1-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.1-1(y_0, a.))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
The TRS R consists of the following rules:
f.1-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)
f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
The set Q consists of the following terms:
f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))
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(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.0-0(h.1(a.), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
F.0-0(h.1(a.), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.1(a.)), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
F.1-0(a., f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.1(a.), x1))
F.1-0(a., f.0-1(f.1-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.1-1(y_0, a.))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
The TRS R consists of the following rules:
f.1-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)
f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
The set Q consists of the following terms:
f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))
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(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.0-0(h.1(a.), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
The TRS R consists of the following rules:
f.1-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)
f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
The set Q consists of the following terms:
f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))
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(h.0(h.1(a.)), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
Strictly oriented rules of the TRS R:
f.1-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)
f.0-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
Used ordering: POLO with Polynomial interpretation [25]:
POL(F.0-0(x1, x2)) = x1 + x2
POL(F.1-0(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(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.1(a.), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
The TRS R consists of the following rules:
f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
The set Q consists of the following terms:
f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))
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(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
The TRS R consists of the following rules:
f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
The set Q consists of the following terms:
f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))
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(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(F.0-0(x1, x2)) = 1 + x1 + x2
POL(F.1-0(x1, x2)) = 1 + x1 + x2
POL(a.) = 0
POL(f.0-0(x1, x2)) = x1 + x2
POL(f.0-1(x1, x2)) = 1 + x1 + x2
POL(f.1-0(x1, x2)) = 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.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)
The set Q consists of the following terms:
f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.