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:
f(X) → if(X, c, f(true))
if(true, X, Y) → X
if(false, X, Y) → Y
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X) → if(X, c, f(true))
if(true, X, Y) → X
if(false, X, Y) → Y
Q is empty.
The TRS is overlay and locally confluent. By [15] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X) → if(X, c, f(true))
if(true, X, Y) → X
if(false, X, Y) → Y
The set Q consists of the following terms:
f(x0)
if(true, x0, x1)
if(false, x0, x1)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(true)
F(X) → IF(X, c, f(true))
The TRS R consists of the following rules:
f(X) → if(X, c, f(true))
if(true, X, Y) → X
if(false, X, Y) → Y
The set Q consists of the following terms:
f(x0)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(true)
F(X) → IF(X, c, f(true))
The TRS R consists of the following rules:
f(X) → if(X, c, f(true))
if(true, X, Y) → X
if(false, X, Y) → Y
The set Q consists of the following terms:
f(x0)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(true)
F(X) → IF(X, c, f(true))
The TRS R consists of the following rules:
f(X) → if(X, c, f(true))
if(true, X, Y) → X
if(false, X, Y) → Y
The set Q consists of the following terms:
f(x0)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(X) → F(true)
The TRS R consists of the following rules:
f(X) → if(X, c, f(true))
if(true, X, Y) → X
if(false, X, Y) → Y
The set Q consists of the following terms:
f(x0)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.