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(s(x)) → s(f(f(p(s(x)))))
f(0) → 0
p(s(x)) → x
Q is empty.
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(x)) → s(f(f(p(s(x)))))
f(0) → 0
p(s(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
f(s(x)) → s(f(f(p(s(x)))))
f(0) → 0
p(s(x)) → x
The set Q is empty.
We have obtained the following QTRS:
s(f(x)) → s(p(f(f(s(x)))))
0'(f(x)) → 0'(x)
s(p(x)) → x
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ RFCMatchBoundsTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
s(f(x)) → s(p(f(f(s(x)))))
0'(f(x)) → 0'(x)
s(p(x)) → x
Q is empty.
Termination of the TRS R could be shown with a Match Bound [6,7] of 2. This implies Q-termination of R.
The following rules were used to construct the certificate:
s(f(x)) → s(p(f(f(s(x)))))
0'(f(x)) → 0'(x)
s(p(x)) → x
The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
93, 94, 95, 96, 98, 97, 99, 100, 102, 101
Node 93 is start node and node 94 is final node.
Those nodes are connect through the following edges:
- 93 to 94 labelled p_1(0), 0'_1(0), f_1(0), s_1(0), p_1(1), f_1(1), s_1(1), 0'_1(1)
- 93 to 95 labelled s_1(0)
- 93 to 99 labelled s_1(1)
- 93 to 97 labelled f_1(1)
- 93 to 101 labelled f_1(2)
- 94 to 94 labelled #_1(0)
- 95 to 96 labelled p_1(0)
- 96 to 97 labelled f_1(0)
- 98 to 94 labelled s_1(0), p_1(1), f_1(1), s_1(1), 0'_1(1)
- 98 to 99 labelled s_1(1)
- 98 to 101 labelled f_1(2)
- 97 to 98 labelled f_1(0)
- 99 to 100 labelled p_1(1)
- 100 to 101 labelled f_1(1)
- 102 to 94 labelled s_1(1), p_1(1), f_1(1), 0'_1(1)
- 102 to 99 labelled s_1(1)
- 102 to 101 labelled f_1(2)
- 101 to 102 labelled f_1(1)