Certification Problem

Input (COPS 1084)

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

+(0,y) y (1)
+(s(x),y) s(+(x,y)) (2)
*(0,y) 0 (3)
*(s(x),y) +(*(x,y),y) (4)
*(x,s(y)) +(x,*(x,y)) (5)

and S contains the following rules:

s(p(x)) x (6)
p(s(x)) x (7)
+(x,0) x (8)
+(x,s(y)) s(+(x,y)) (9)
+(x,p(y)) p(+(x,y)) (10)
-(x,0) x (11)
-(x,s(y)) p(-(x,y)) (12)
-(x,p(y)) s(-(x,y)) (13)
*(x,0) 0 (14)
*(x,s(y)) +(*(x,y),x) (15)
*(x,p(y)) -(*(x,y),x) (16)

The underlying signature is as follows:

{*/2, +/2, -/2, 0/0, p/1, s/1}

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 = *(0,p(y))
S -(*(0,y),0)
= s1
and
t0 = *(0,p(y))
R 0
= t1
There is no possibility to join s1R*·←S* t1 for the following reason: