YES Termination w.r.t. Q proof of SK90_4.37.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:

a(b(a(x))) → b(a(x))

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:

a(b(a(x))) → b(a(x))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

2, 4, 5, 7

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

Those nodes are connected through the following edges:

  • 2 to 5 labelled b_1(0)
  • 4 to 4 labelled #_1(0)
  • 5 to 4 labelled a_1(0)
  • 5 to 7 labelled b_1(1)
  • 7 to 4 labelled a_1(1)
  • 7 to 7 labelled b_1(1)

(2) YES