YES Termination w.r.t. Q proof of SK90_2.47.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:

a(b(x)) → b(a(x))
a(c(x)) → 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:

b(a(x)) → a(b(x))
c(a(x)) → x

Q is empty.

(3) 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:

b(a(x)) → a(b(x))
c(a(x)) → x

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

1, 3, 5, 8

Node 1 is start node and node 3 is final node.

Those nodes are connected through the following edges:

  • 1 to 5 labelled a_1(0)
  • 1 to 3 labelled b_1(0), a_1(0), c_1(0), b_1(1), a_1(1), c_1(1)
  • 1 to 8 labelled a_1(1)
  • 1 to 1 labelled a_1(1)
  • 3 to 3 labelled #_1(0)
  • 5 to 3 labelled b_1(0)
  • 5 to 1 labelled a_1(1)
  • 8 to 3 labelled b_1(1)
  • 8 to 1 labelled a_1(1)

(4) YES