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:
or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
MEM2(x, union2(y, z)) -> MEM2(x, z)
MEM2(x, union2(y, z)) -> OR2(mem2(x, y), mem2(x, z))
MEM2(x, union2(y, z)) -> MEM2(x, y)
The TRS R consists of the following rules:
or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MEM2(x, union2(y, z)) -> MEM2(x, z)
MEM2(x, union2(y, z)) -> OR2(mem2(x, y), mem2(x, z))
MEM2(x, union2(y, z)) -> MEM2(x, y)
The TRS R consists of the following rules:
or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MEM2(x, union2(y, z)) -> MEM2(x, z)
MEM2(x, union2(y, z)) -> MEM2(x, y)
The TRS R consists of the following rules:
or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))
Q is empty.
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.
MEM2(x, union2(y, z)) -> MEM2(x, z)
MEM2(x, union2(y, z)) -> MEM2(x, y)
Used argument filtering: MEM2(x1, x2) = x2
union2(x1, x2) = union2(x1, x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
or2(true, y) -> true
or2(x, true) -> true
or2(false, false) -> false
mem2(x, nil) -> false
mem2(x, set1(y)) -> =2(x, y)
mem2(x, union2(y, z)) -> or2(mem2(x, y), mem2(x, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.