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:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
The set Q consists of the following terms:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, x), z))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), y)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), y)), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, app2(app2(:, x), y)), z))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, x), y))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, x)
The TRS R consists of the following rules:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
The set Q consists of the following terms:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, x), z))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), y)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), y)), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, app2(app2(:, x), y)), z))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, app2(app2(:, x), y))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(:, x)
The TRS R consists of the following rules:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
The set Q consists of the following terms:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 4 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), y)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), y)), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u)
The TRS R consists of the following rules:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
The set Q consists of the following terms:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), y)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, x), y)), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, x), z)
APP2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> APP2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u)
Used argument filtering: APP2(x1, x2) = x1
app2(x1, x2) = app2(x1, x2)
: = :
C = C
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x)), y)), z)), u) -> app2(app2(:, app2(app2(:, x), z)), app2(app2(:, app2(app2(:, app2(app2(:, x), y)), z)), u))
The set Q consists of the following terms:
app2(app2(:, app2(app2(:, app2(app2(:, app2(app2(:, C), x0)), x1)), x2)), x3)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.