Certification Problem
Input (COPS 1081)
We consider two TRSs R and S where R contains the rules
f(g(f(x))) |
→ |
g(f(g(x))) |
(1) |
f(c) |
→ |
c |
(2) |
g(c) |
→ |
c |
(3) |
and S contains the following rules:
g(x) |
→ |
h(k(x)) |
(4) |
g(x) |
→ |
x |
(5) |
h(k(x)) |
→ |
f(x) |
(6) |
f(x) |
→ |
x |
(7) |
k(c) |
→ |
c |
(8) |
f(c) |
→ |
g(c) |
(9) |
The underlying signature is as follows:
{c/0, f/1, g/1, h/1, k/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
|
= |
g(c) |
|
→S
|
h(k(c)) |
|
= |
s1
|
and
There is no possibility to join
s1→R*·←S*
t1
for the following reason:
- When applying the cap-function on both terms (where variables may be treated like constants)
then the resulting terms do not unify.