Certification Problem

Input (COPS 1122)

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

if(true,a,x) a (1)
if(true,b,x) b (2)
if(true,g(a),x) g(a) (3)
if(true,g(b),x) g(b) (4)
if(false,x,a) a (5)
if(false,x,b) b (6)
if(false,x,g(a)) g(a) (7)
if(false,x,g(b)) g(b) (8)
g(a) g(g(a)) (9)
g(b) a (10)
f'(a,b) b (11)
f'(g(g(a)),x) b (12)

and S contains the following rules:

h(c,c) h(b,f(b)) (13)
f(h(a,h(c,a))) c (14)
h(c,b) f(a) (15)

The underlying signature is as follows:

{a/0, b/0, c/0, f/1, g/1, h/2, f'/2, if/3, true/0, false/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).