YES Termination w.r.t. Q proof of Strategy_removed_AG01_4.20a.ari

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 RFCMatchBoundsTRSProof (⇔, 0 ms)
↳2 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(f(x)) → f(x)
f(s(x)) → f(x)
g(s(0)) → g(f(s(0)))

Q is empty.

(1) RFCMatchBoundsTRSProof (EQUIVALENT transformation)

Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R.
The following rules were used to construct the certificate:

f(f(x)) → f(x)
f(s(x)) → f(x)
g(s(0)) → g(f(s(0)))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

2, 3, 5, 6, 7

Node 2 is start node and node 3 is final node.

Those nodes are connected through the following edges:

  • 2 to 3 labelled f_1(0), f_1(1)
  • 2 to 5 labelled g_1(0)
  • 3 to 3 labelled #_1(0)
  • 5 to 6 labelled f_1(0)
  • 5 to 7 labelled f_1(1)
  • 6 to 7 labelled s_1(0)
  • 7 to 3 labelled 0(0)

(2) YES