Certification Problem

Input (COPS 47)

We consider the TRS containing the following rules:

F(x,x)	→	A	(1)
G(x)	→	F(x,G(x))	(2)
C	→	G(C)	(3)

The underlying signature is as follows:

{F/2, A/0, G/1, C/0}

Property / Task

Prove or disprove confluence.

Answer / Result

No.

Proof (by csi @ CoCo 2023)

1 Redundant Rules Transformation

To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:

F(x,x)	→	A	(1)
G(x)	→	F(x,G(x))	(2)
C	→	G(C)	(3)
G(C)	→	F(G(C),G(C))	(4)
G(C)	→	F(C,G(G(C)))	(5)
C	→	F(C,G(C))	(6)
C	→	G(G(C))	(7)
F(F(x,x),A)	→	A	(8)
F(A,F(x,x))	→	A	(9)
F(C,G(C))	→	A	(10)
F(G(C),C)	→	A	(11)

All redundant rules that were added or removed can be simulated in 3 steps .

1.1 Non-Joinable Fork

The system is not confluent due to the following forking derivations.

t₀	=	G(C)
	→	G(F(C,G(C)))
	→	G(A)
	=	t₂

t₀	=	G(C)
	→	F(G(C),G(C))
	→	A
	=	t₂

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:
    
    {1028}
  - transitions:
    
    A → 1029
    
    F(1029,1028) → 1028
    
    G(1029) → 1028
  The automaton is closed under rewriting as it is state-compatible w.r.t. the following relation.
  
  1029 » 1029
  
  1028 » 1028
- Automaton 2
  - final states:
    
    {7}
  - transitions:
    
    A → 7
  The automaton is closed under rewriting as it is state-compatible w.r.t. the following relation.
  7 » 7