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(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)
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(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)
The set Q consists of the following terms:
app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(app2(curry, f), x), y) -> APP2(app2(f, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(app2(curry, f), x), y) -> APP2(f, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
ADD -> APP2(curry, plus)
The TRS R consists of the following rules:
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)
The set Q consists of the following terms:
app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add
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(curry, f), x), y) -> APP2(app2(f, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(app2(curry, f), x), y) -> APP2(f, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
ADD -> APP2(curry, plus)
The TRS R consists of the following rules:
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)
The set Q consists of the following terms:
app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
The TRS R consists of the following rules:
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)
The set Q consists of the following terms:
app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add
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(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
Used argument filtering: APP2(x1, x2) = x1
app2(x1, x2) = app1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)
The set Q consists of the following terms:
app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(app2(curry, f), x), y) -> APP2(app2(f, x), y)
APP2(app2(app2(curry, f), x), y) -> APP2(f, x)
The TRS R consists of the following rules:
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)
The set Q consists of the following terms:
app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add
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(curry, f), x), y) -> APP2(app2(f, x), y)
APP2(app2(app2(curry, f), x), y) -> APP2(f, x)
Used argument filtering: APP2(x1, x2) = x1
app2(x1, x2) = app2(x1, x2)
curry = curry
plus = plus
0 = 0
s = s
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(app2(curry, f), x), y) -> app2(app2(f, x), y)
add -> app2(curry, plus)
The set Q consists of the following terms:
app2(app2(plus, 0), x0)
app2(app2(plus, app2(s, x0)), x1)
app2(app2(app2(curry, x0), x1), x2)
add
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.