Certification Problem

Input (COPS 497)

We consider the TRS containing the following rules:

d(x,x)	→	0	(1)
f(x)	→	d(x,f(x))	(2)
c	→	f(c)	(3)

The underlying signature is as follows:

{d/2, 0/0, f/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:

d(x,x)	→	0	(1)
f(x)	→	d(x,f(x))	(2)
c	→	f(c)	(3)
f(c)	→	d(f(c),f(c))	(4)
f(c)	→	d(c,f(f(c)))	(5)
c	→	d(c,f(c))	(6)
c	→	f(f(c))	(7)
d(d(x,x),0)	→	0	(8)
d(0,d(x,x))	→	0	(9)
d(c,f(c))	→	0	(10)
d(f(c),c)	→	0	(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	=	f(c)
	→	f(d(c,f(c)))
	→	f(0)
	=	t₂

t₀	=	f(c)
	→	d(f(c),f(c))
	→	0
	=	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:
    
    0 → 1029
    
    d(1029,1028) → 1028
    
    f(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:
    
    0 → 7
  The automaton is closed under rewriting as it is state-compatible w.r.t. the following relation.
  7 » 7