Certification Problem

Input (COPS 690)

We consider the TRS containing the following rules:

h(f(f(c)),b)	→	f(h(h(h(c,h(f(h(c,f(b))),a)),b),c))	(1)
c	→	c	(2)
f(f(h(h(f(a),a),c)))	→	f(h(f(c),b))	(3)
h(f(h(f(b),h(h(f(h(c,f(c))),b),a))),h(a,c))	→	c	(4)

The underlying signature is as follows:

{h/2, f/1, c/0, b/0, a/0}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by Hakusan @ CoCo 2023)

1 Compositional Rule Labeling with Parallel Critical Pairs

Confluence is proven by compositional rule labeling with parallel critical pairs.
The following labeling functions phi and psi are used (if only one is displayed, then phi = psi).

- h(f(f(c)),b) → f(h(h(h(c,h(f(h(c,f(b))),a)),b),c)) (1)
  ↦ 1
- c → c (2)
  ↦ 1
- f(f(h(h(f(a),a),c))) → f(h(f(c),b)) (3)
  ↦ 1
- h(f(h(f(b),h(h(f(h(c,f(c))),b),a))),h(a,c)) → c (4)
  ↦ 1
- h(f(f(c)),b) → f(h(h(h(c,h(f(h(c,f(b))),a)),b),c)) (1)
  ↦ 1
- c → c (2)
  ↦ 1
- f(f(h(h(f(a),a),c))) → f(h(f(c),b)) (3)
  ↦ 1
- h(f(h(f(b),h(h(f(h(c,f(c))),b),a))),h(a,c)) → c (4)
  ↦ 1

The non-0-0 parallel critical pairs are joined as follows.

The remaining rules are handled recursively.

There are no rules.

1.1 (Weakly) Orthogonal

Confluence is proven since the TRS is (weakly) orthogonal.