Certification Problem

Input (COPS 95)

We consider the TRS containing the following rules:

W(B(x))	→	I(x)	(1)
B(S(x))	→	S(x)	(2)
W(x)	→	I(x)	(3)

The underlying signature is as follows:

{W/1, B/1, I/1, S/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	=	W(B(f6))
	→	I(B(f6))
	=	t₁

t₀	=	W(B(f6))
	→	I(f6)
	=	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:
    
    I(3) → 1
    
    B(2) → 3
    
    f6 → 2
  The automaton is closed under rewriting as it is compatible.
- Automaton 2
  - final states:
    
    {4}
  - transitions:
    
    I(5) → 4
    
    f6 → 5
  The automaton is closed under rewriting as it is compatible.