Certification Problem
Input (COPS 10)
We consider the TRS containing the following rules:
|
f(g(f(x))) |
→ |
x |
(1) |
|
f(g(x)) |
→ |
g(f(x)) |
(2) |
The underlying signature is as follows:
{f/1, g/1}Property / Task
Prove or disprove confluence.Answer / Result
No.Proof (by csi @ CoCo 2023)
1 Non-Joinable Fork
The system is not confluent due to the following forking derivations.
| t0
|
= |
f(g(f(f4))) |
|
→
|
g(f(f(f4))) |
|
= |
t1
|
| t0
|
= |
f(g(f(f4))) |
|
→
|
f4 |
|
= |
t1
|
The two resulting terms cannot be joined for the following reason:
-
The reachable terms of these two terms are approximated via the following two tree automata,
and the tree automata have an empty intersection.
-
Automaton 1
-
final states:
{8}
-
transitions:
| f4 |
→ |
9 |
|
g(11) |
→ |
8 |
|
f(9) |
→ |
10 |
|
f(10) |
→ |
11 |
The automaton is closed under rewriting as it is compatible.
-
Automaton 2
-
final states:
{12}
-
transitions:
The automaton is closed under rewriting as it is compatible.