Certification Problem
Input (COPS 1059)
We consider two TRSs R and S where R contains the rules
| a |
→ |
f(a) |
(1) |
| a |
→ |
g(h(a)) |
(2) |
|
f(x) |
→ |
h(g(x)) |
(3) |
|
h(x) |
→ |
f(g(x)) |
(4) |
and S contains the following rules:
| a |
→ |
b |
(5) |
| a |
→ |
c |
(6) |
| a |
→ |
e |
(7) |
| b |
→ |
d |
(8) |
| c |
→ |
a |
(9) |
| d |
→ |
a |
(10) |
| d |
→ |
e |
(11) |
|
g(x) |
→ |
h(a) |
(12) |
|
h(x) |
→ |
e |
(13) |
The underlying signature is as follows:
{a/0, b/0, c/0, d/0, e/0, f/1, g/1, h/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.
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.