YES Termination w.r.t. Q proof of AG01_3.33.ari

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

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

(0) Obligation:

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

p(f(f(x))) → q(f(g(x)))
p(g(g(x))) → q(g(f(x)))
q(f(f(x))) → p(f(g(x)))
q(g(g(x))) → p(g(f(x)))

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

f(f(p(x))) → g(f(q(x)))
g(g(p(x))) → f(g(q(x)))
f(f(q(x))) → g(f(p(x)))
g(g(q(x))) → f(g(p(x)))

Q is empty.

(3) RFCMatchBoundsTRSProof (EQUIVALENT transformation)

Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 0. This implies Q-termination of R.
The following rules were used to construct the certificate:

f(f(p(x))) → g(f(q(x)))
g(g(p(x))) → f(g(q(x)))
f(f(q(x))) → g(f(p(x)))
g(g(q(x))) → f(g(p(x)))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

2, 4, 9, 10, 11, 12

Node 2 is start node and node 4 is final node.

Those nodes are connected through the following edges:

  • 2 to 9 labelled g_1(0)
  • 2 to 11 labelled f_1(0)
  • 4 to 4 labelled #_1(0)
  • 9 to 10 labelled f_1(0)
  • 10 to 4 labelled q_1(0), p_1(0)
  • 11 to 12 labelled g_1(0)
  • 12 to 4 labelled q_1(0), p_1(0)

(4) YES