Certification Problem

Input (COPS 1101)

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

b → h(b,a) (1)
b → h(f(b),f(h(b,a))) (2)

and S contains the following rules:

f(b) → a (3)
f(b) → f(c) (4)
f(c) → f(b) (5)
f(c) → d (6)
b → e (7)
c → e' (8)
f(e) → a (9)
f(e') → d (10)

The underlying signature is as follows:

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

Property / Task

Prove or disprove commutation.

Answer / Result

No.

Proof (by ACP @ CoCo 2023)

1 Non-Joinable Fork

The systems are not commuting due to the following forking derivations.
t0 = b
→S e
= s1
and
t0 = b
→R h(b,a)
= t1
There is no possibility to join s1→R*·←S* t1 for the following reason: