Certification Problem
Input (COPS 98)
We consider the TRS containing the following rules:
|
I(x) |
→ |
I(B(x)) |
(1) |
|
F(E(x),x) |
→ |
G(x) |
(2) |
|
E(x) |
→ |
x |
(3) |
The underlying signature is as follows:
{I/1, B/1, F/2, E/1, G/1}Property / Task
Prove or disprove confluence.Answer / Result
No.Proof (by csi @ CoCo 2023)
1 Persistent Decomposition (Many-Sorted)
Non-confluence
is proven, because
a
system
induced by the sorts in the following many-sorted sort attachment
is not
confluent.
| I |
: |
3 → 0 |
| B |
: |
3 → 3 |
| F |
: |
2 ⨯ 2 → 1 |
| E |
: |
2 → 2 |
| G |
: |
2 → 1 |
The subsystem is(1.1)
|
F(E(x),x) |
→ |
G(x) |
(2) |
|
E(x) |
→ |
x |
(3) |
1.1 Non-Joinable Fork
The system is not confluent due to the following forking derivations.
| t0
|
= |
F(E(f5),f5) |
|
→
|
F(f5,f5) |
|
= |
t1
|
| t0
|
= |
F(E(f5),f5) |
|
→
|
G(f5) |
|
= |
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:
{1}
-
transitions:
The automaton is closed under rewriting as it is state-compatible w.r.t. the following relation.
-
Automaton 2
-
final states:
{4}
-
transitions:
The automaton is closed under rewriting as it is state-compatible w.r.t. the following relation.