Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))
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:
del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))
The set Q consists of the following terms:
del1(.2(x0, .2(x1, x2)))
f4(true, x0, x1, x2)
f4(false, x0, x1, x2)
=2(nil, nil)
=2(.2(x0, x1), nil)
=2(nil, .2(x0, x1))
=2(.2(x0, x1), .2(u, v))
Q DP problem:
The TRS P consists of the following rules:
F4(false, x, y, z) -> DEL1(.2(y, z))
=12(.2(x, y), .2(u, v)) -> =12(x, u)
F4(true, x, y, z) -> DEL1(.2(y, z))
DEL1(.2(x, .2(y, z))) -> F4(=2(x, y), x, y, z)
DEL1(.2(x, .2(y, z))) -> =12(x, y)
=12(.2(x, y), .2(u, v)) -> =12(y, v)
The TRS R consists of the following rules:
del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))
The set Q consists of the following terms:
del1(.2(x0, .2(x1, x2)))
f4(true, x0, x1, x2)
f4(false, x0, x1, x2)
=2(nil, nil)
=2(.2(x0, x1), nil)
=2(nil, .2(x0, x1))
=2(.2(x0, x1), .2(u, v))
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:
F4(false, x, y, z) -> DEL1(.2(y, z))
=12(.2(x, y), .2(u, v)) -> =12(x, u)
F4(true, x, y, z) -> DEL1(.2(y, z))
DEL1(.2(x, .2(y, z))) -> F4(=2(x, y), x, y, z)
DEL1(.2(x, .2(y, z))) -> =12(x, y)
=12(.2(x, y), .2(u, v)) -> =12(y, v)
The TRS R consists of the following rules:
del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))
The set Q consists of the following terms:
del1(.2(x0, .2(x1, x2)))
f4(true, x0, x1, x2)
f4(false, x0, x1, x2)
=2(nil, nil)
=2(.2(x0, x1), nil)
=2(nil, .2(x0, x1))
=2(.2(x0, x1), .2(u, v))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F4(false, x, y, z) -> DEL1(.2(y, z))
F4(true, x, y, z) -> DEL1(.2(y, z))
DEL1(.2(x, .2(y, z))) -> F4(=2(x, y), x, y, z)
The TRS R consists of the following rules:
del1(.2(x, .2(y, z))) -> f4(=2(x, y), x, y, z)
f4(true, x, y, z) -> del1(.2(y, z))
f4(false, x, y, z) -> .2(x, del1(.2(y, z)))
=2(nil, nil) -> true
=2(.2(x, y), nil) -> false
=2(nil, .2(y, z)) -> false
=2(.2(x, y), .2(u, v)) -> and2(=2(x, u), =2(y, v))
The set Q consists of the following terms:
del1(.2(x0, .2(x1, x2)))
f4(true, x0, x1, x2)
f4(false, x0, x1, x2)
=2(nil, nil)
=2(.2(x0, x1), nil)
=2(nil, .2(x0, x1))
=2(.2(x0, x1), .2(u, v))
We have to consider all minimal (P,Q,R)-chains.