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
↳ Overlay + Local Confluence
↳ 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.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ 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
The set Q consists of the following terms:
f(s(x0))
f(0)
p(s(x0))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(f(p(s(x))))
F(s(x)) → F(p(s(x)))
F(s(x)) → P(s(x))
The TRS R consists of the following rules:
f(s(x)) → s(f(f(p(s(x)))))
f(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
f(s(x0))
f(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(f(p(s(x))))
F(s(x)) → F(p(s(x)))
F(s(x)) → P(s(x))
The TRS R consists of the following rules:
f(s(x)) → s(f(f(p(s(x)))))
f(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
f(s(x0))
f(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(f(p(s(x))))
F(s(x)) → F(p(s(x)))
The TRS R consists of the following rules:
f(s(x)) → s(f(f(p(s(x)))))
f(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
f(s(x0))
f(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
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
↳ Overlay + Local Confluence
↳ 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:
101, 102, 103, 104, 106, 105, 107, 108, 110, 109
Node 101 is start node and node 102 is final node.
Those nodes are connect through the following edges:
- 101 to 102 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)
- 101 to 103 labelled s_1(0)
- 101 to 107 labelled s_1(1)
- 101 to 105 labelled f_1(1)
- 101 to 109 labelled f_1(2)
- 102 to 102 labelled #_1(0)
- 103 to 104 labelled p_1(0)
- 104 to 105 labelled f_1(0)
- 106 to 102 labelled s_1(0), p_1(1), f_1(1), s_1(1), 0'_1(1)
- 106 to 107 labelled s_1(1)
- 106 to 109 labelled f_1(2)
- 105 to 106 labelled f_1(0)
- 107 to 108 labelled p_1(1)
- 108 to 109 labelled f_1(1)
- 110 to 102 labelled s_1(1), p_1(1), f_1(1), 0'_1(1)
- 110 to 107 labelled s_1(1)
- 110 to 109 labelled f_1(2)
- 109 to 110 labelled f_1(1)