Certification Problem

Input (COPS 949)

We consider the TRS containing the following rules:

0(0(0(1(2(1(1(1(0(0(1(0(1(x)))))))))))))	→	0(0(1(0(1(1(0(2(0(0(2(0(1(0(1(0(1(x)))))))))))))))))	(1)
0(0(1(0(0(1(1(1(0(1(2(0(1(x)))))))))))))	→	0(1(1(0(0(1(0(1(1(1(1(0(2(0(1(1(1(x)))))))))))))))))	(2)
0(0(1(0(1(0(1(0(0(1(0(2(0(x)))))))))))))	→	0(1(1(1(0(1(1(2(1(2(0(1(1(1(0(1(1(x)))))))))))))))))	(3)
0(0(1(1(1(1(1(2(1(1(2(1(1(x)))))))))))))	→	0(1(2(0(1(0(0(1(2(1(0(0(1(0(1(1(1(x)))))))))))))))))	(4)
0(1(0(2(0(2(1(0(0(1(0(1(1(x)))))))))))))	→	0(0(0(1(1(1(1(1(0(2(2(0(1(1(1(0(1(x)))))))))))))))))	(5)
0(1(0(2(2(1(1(2(1(2(2(0(1(x)))))))))))))	→	0(1(2(2(0(0(1(0(1(0(2(1(0(2(2(0(1(x)))))))))))))))))	(6)
0(1(1(0(1(1(0(1(0(2(1(0(0(x)))))))))))))	→	0(1(2(0(1(1(0(1(1(1(1(1(0(0(1(0(0(x)))))))))))))))))	(7)
0(1(1(1(2(1(2(0(1(2(1(0(1(x)))))))))))))	→	0(1(0(1(0(1(1(0(0(1(0(2(0(1(0(0(1(x)))))))))))))))))	(8)
0(2(0(0(1(1(2(0(1(0(1(0(2(x)))))))))))))	→	0(0(1(2(0(0(1(0(1(0(1(1(0(0(1(1(1(x)))))))))))))))))	(9)
1(0(2(0(0(2(0(1(2(0(1(0(1(x)))))))))))))	→	1(1(0(0(1(0(2(1(2(0(1(1(1(0(1(1(1(x)))))))))))))))))	(10)
1(0(2(0(2(1(0(2(0(1(1(2(0(x)))))))))))))	→	1(2(0(2(0(0(1(0(1(0(0(0(1(2(0(1(0(x)))))))))))))))))	(11)
1(1(0(0(2(2(2(0(1(2(0(1(1(x)))))))))))))	→	1(0(1(1(1(0(0(1(2(0(1(2(0(0(0(0(1(x)))))))))))))))))	(12)
1(2(0(1(0(2(0(1(0(1(2(1(0(x)))))))))))))	→	1(0(1(1(2(0(1(0(0(1(0(0(1(0(1(2(0(x)))))))))))))))))	(13)
1(2(1(2(0(0(0(1(1(1(0(0(1(x)))))))))))))	→	1(1(1(1(0(2(0(1(1(0(1(1(2(2(0(0(1(x)))))))))))))))))	(14)
2(0(0(0(2(2(0(2(2(0(1(0(1(x)))))))))))))	→	2(0(0(1(1(2(1(1(2(0(0(1(2(1(2(0(1(x)))))))))))))))))	(15)
2(0(2(1(0(0(1(0(0(0(1(1(1(x)))))))))))))	→	0(2(1(2(1(0(1(1(1(0(1(0(1(0(0(1(1(x)))))))))))))))))	(16)
2(0(2(1(1(1(1(0(2(0(1(0(1(x)))))))))))))	→	2(1(2(0(2(0(1(0(1(2(0(1(0(1(1(1(1(x)))))))))))))))))	(17)
2(1(2(2(1(1(2(2(0(1(1(0(1(x)))))))))))))	→	2(1(1(1(2(0(1(2(2(0(0(0(1(1(1(0(1(x)))))))))))))))))	(18)
2(2(1(1(1(1(0(1(1(2(0(1(0(x)))))))))))))	→	0(1(0(1(1(1(1(1(2(2(0(1(2(1(1(1(0(x)))))))))))))))))	(19)

The underlying signature is as follows:

{0/1, 1/1, 2/1}

Property / Task

Prove or disprove confluence.

Answer / Result

No.

Proof (by ACP @ CoCo 2023)

1 Non-Joinable Fork

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

t₀	=	0(0(0(1(2(1(1(1(0(0(1(0(1(0(1(0(0(1(0(2(0(c_1)))))))))))))))))))))
	→	0(0(1(0(1(1(0(2(0(0(2(0(1(0(1(0(1(0(1(0(0(1(0(2(0(c_1)))))))))))))))))))))))))
	=	t₁

t₀	=	0(0(0(1(2(1(1(1(0(0(1(0(1(0(1(0(0(1(0(2(0(c_1)))))))))))))))))))))
	→	0(0(0(1(2(1(1(1(0(1(1(1(0(1(1(2(1(2(0(1(1(1(0(1(1(c_1)))))))))))))))))))))))))
	=	t₁

The two resulting terms cannot be joined for the following reason:

We take the usable rules of the first term (w.r.t. the TRS for the first term) and the usable rules of the second term (w.r.t. the TRS for the second term). Then the terms are not joinable w.r.t. the resulting TRSs.
We filter all terms and rules w.r.t. the following argument filter.

π(0) = [1]

π(1) = 1

π(2) = [1]

π(c_1) = []

Then the resulting terms are not joinable w.r.t. the resulting TRSs.
We take the usable rules of the first term (w.r.t. the TRS for the first term) and the usable rules of the second term (w.r.t. the TRS for the second term). Then the terms are not joinable w.r.t. the resulting TRSs.
The first mentioned term is strictly larger than the second one (or vice versa). Here, the following discrimination pair has been used w.r.t. the following interpretation. Moreover, the (reversed) rules are weakly decreasing. The discrimination pair is given by a

π(0)	=	[1]
π(1)	=	1
π(2)	=	[1]
π(c_1)	=	[]