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:
merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))
MERGE2(.2(x, y), .2(u, v)) -> IF3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(.2(x, y), v)
++12(.2(x, y), z) -> ++12(y, z)
The TRS R consists of the following rules:
merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x
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:
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))
MERGE2(.2(x, y), .2(u, v)) -> IF3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(.2(x, y), v)
++12(.2(x, y), z) -> ++12(y, z)
The TRS R consists of the following rules:
merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
++12(.2(x, y), z) -> ++12(y, z)
The TRS R consists of the following rules:
merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x
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.
++12(.2(x, y), z) -> ++12(y, z)
Used argument filtering: ++12(x1, x2) = x1
.2(x1, x2) = .1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x
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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(.2(x, y), v)
The TRS R consists of the following rules:
merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x
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.
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(.2(x, y), v)
Used argument filtering: MERGE2(x1, x2) = x2
.2(x1, x2) = .1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))
The TRS R consists of the following rules:
merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x
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.
MERGE2(.2(x, y), .2(u, v)) -> MERGE2(y, .2(u, v))
Used argument filtering: MERGE2(x1, x2) = x1
.2(x1, x2) = .1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
merge2(nil, y) -> y
merge2(x, nil) -> x
merge2(.2(x, y), .2(u, v)) -> if3(<2(x, u), .2(x, merge2(y, .2(u, v))), .2(u, merge2(.2(x, y), v)))
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))
if3(true, x, y) -> x
if3(false, x, y) -> x
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.