YES Termination w.r.t. Q proof of AG01_3.7.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:

half(0) → 0
half(s(s(x))) → s(half(x))
log(s(0)) → 0
log(s(s(x))) → s(log(s(half(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:

0'(half(x)) → 0'(x)
s(s(half(x))) → half(s(x))
0'(s(log(x))) → 0'(x)
s(s(log(x))) → half(s(log(s(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:

0'(half(x)) → 0'(x)
s(s(half(x))) → half(s(x))
0'(s(log(x))) → 0'(x)
s(s(log(x))) → half(s(log(s(x))))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

1, 4, 6, 10, 11, 12, 14, 15, 16, 17

Node 1 is start node and node 4 is final node.

Those nodes are connected through the following edges:

  • 1 to 4 labelled 0'_1(0), 0'_1(1)
  • 1 to 6 labelled half_1(0)
  • 1 to 10 labelled half_1(0)
  • 4 to 4 labelled #_1(0)
  • 6 to 4 labelled s_1(0)
  • 6 to 14 labelled half_1(1)
  • 6 to 15 labelled half_1(1)
  • 10 to 11 labelled s_1(0)
  • 11 to 12 labelled log_1(0)
  • 12 to 4 labelled s_1(0)
  • 12 to 14 labelled half_1(1)
  • 12 to 15 labelled half_1(1)
  • 14 to 4 labelled s_1(1)
  • 14 to 14 labelled half_1(1)
  • 14 to 15 labelled half_1(1)
  • 15 to 16 labelled s_1(1)
  • 16 to 17 labelled log_1(1)
  • 17 to 4 labelled s_1(1)
  • 17 to 14 labelled half_1(1)
  • 17 to 15 labelled half_1(1)

(4) YES