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(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
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(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
The set Q consists of the following terms:
app2(app2(mapbt, x0), app2(leaf, x1))
app2(app2(mapbt, x0), app2(app2(app2(branch, x1), x2), x3))
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), r)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), l)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(branch, app2(f, x))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(leaf, app2(f, x))
The TRS R consists of the following rules:
app2(app2(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
The set Q consists of the following terms:
app2(app2(mapbt, x0), app2(leaf, x1))
app2(app2(mapbt, x0), app2(app2(app2(branch, 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(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), r)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), l)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(branch, app2(f, x))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(leaf, app2(f, x))
The TRS R consists of the following rules:
app2(app2(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
The set Q consists of the following terms:
app2(app2(mapbt, x0), app2(leaf, x1))
app2(app2(mapbt, x0), app2(app2(app2(branch, 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(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), r)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), l)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(f, x)
The TRS R consists of the following rules:
app2(app2(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
The set Q consists of the following terms:
app2(app2(mapbt, x0), app2(leaf, x1))
app2(app2(mapbt, x0), app2(app2(app2(branch, 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(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), r)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), l)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(f, x)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app2(x1, x2)
branch = branch
leaf = leaf
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(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
The set Q consists of the following terms:
app2(app2(mapbt, x0), app2(leaf, x1))
app2(app2(mapbt, x0), app2(app2(app2(branch, 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.