Certification Problem

Input (COPS 1108)

We consider two TRSs R and S where R contains the rules

b → f(a) (1)
a → c (2)
a → a (3)
h(c,f(f(f(a)))) → h(f(f(c)),h(b,a)) (4)
f(f(f(f(h(h(c,h(f(c),f(f(h(h(h(h(f(a),a),c),a),h(h(h(b,a),f(h(f(h(f(b),f(f(b)))),a))),h(c,f(h(b,f(f(h(h(c,f(a)),a)))))))))))),f(f(b))))))) → c (5)

and S contains the following rules:

a1 → b1 (6)
a1 → c1 (7)
b1 → b2 (8)
c1 → c2 (9)
a2 → b2 (10)
a2 → c2 (11)
b2 → b3 (12)
c2 → c3 (13)
a3 → b3 (14)
a3 → c3 (15)
b3 → b4 (16)
c3 → c4 (17)
a4 → b4 (18)
a4 → c4 (19)
b4 → b5 (20)
c4 → c5 (21)
a5 → b5 (22)
a5 → c5 (23)
b5 → b6 (24)
c5 → c6 (25)
a6 → b6 (26)
a6 → c6 (27)
b6 → b7 (28)
c6 → b7 (29)
b7 → b1 (30)
b7 → c1 (31)

The underlying signature is as follows:

{a/0, b/0, c/0, f/1, h/2, a1/0, a2/0, a3/0, a4/0, a5/0, a6/0, b1/0, b2/0, b3/0, b4/0, b5/0, b6/0, b7/0, c1/0, c2/0, c3/0, c4/0, c5/0, c6/0}

Property / Task

Prove or disprove commutation.

Answer / Result

Yes.

Proof (by ACP @ CoCo 2023)

1 Development Closed

Commutation is proven since the TRSs are development closed. The joins can be performed using 5 step(s).