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:
f(g(X)) → g(f(f(X)))
f(h(X)) → h(g(X))
Q is empty.
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(X)) → g(f(f(X)))
f(h(X)) → h(g(X))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
f(g(X)) → g(f(f(X)))
f(h(X)) → h(g(X))
The set Q is empty.
We have obtained the following QTRS:
g(f(x)) → f(f(g(x)))
h(f(x)) → g(h(x))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ RFCMatchBoundsTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(f(x)) → f(f(g(x)))
h(f(x)) → g(h(x))
Q is empty.
Termination of the TRS R could be shown with a Match Bound [6,7] of 1. This implies Q-termination of R.
The following rules were used to construct the certificate:
g(f(x)) → f(f(g(x)))
h(f(x)) → g(h(x))
The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
129, 130, 132, 131, 133, 134, 136, 135
Node 129 is start node and node 130 is final node.
Those nodes are connect through the following edges:
- 129 to 131 labelled f_1(0)
- 129 to 133 labelled g_1(0)
- 130 to 130 labelled #_1(0)
- 132 to 130 labelled g_1(0)
- 132 to 135 labelled f_1(1)
- 131 to 132 labelled f_1(0)
- 133 to 130 labelled h_1(0)
- 133 to 134 labelled g_1(1)
- 134 to 130 labelled h_1(1)
- 134 to 134 labelled g_1(1)
- 136 to 130 labelled g_1(1)
- 136 to 135 labelled f_1(1)
- 135 to 136 labelled f_1(1)