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:
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
IF(if(x, y, z), u, v) → IF(z, u, v)
IF(if(x, y, z), u, v) → IF(x, if(y, u, v), if(z, u, v))
IF(if(x, y, z), u, v) → IF(y, u, v)
The TRS R consists of the following rules:
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
IF(if(x, y, z), u, v) → IF(z, u, v)
IF(if(x, y, z), u, v) → IF(x, if(y, u, v), if(z, u, v))
IF(if(x, y, z), u, v) → IF(y, u, v)
The TRS R consists of the following rules:
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
IF(if(x, y, z), u, v) → IF(z, u, v)
IF(if(x, y, z), u, v) → IF(x, if(y, u, v), if(z, u, v))
IF(if(x, y, z), u, v) → IF(y, u, v)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3) = x1
if(x1, x2, x3) = if(x1, x2, x3)
Lexicographic Path Order [19].
Precedence:
trivial
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
if(if(x, y, z), u, v) → if(x, if(y, u, v), if(z, u, v))
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.