Certification Problem

Input (COPS 1002)

We consider the TRS containing the following rules:

c(a(x)) b(c(x)) (1)
c(b(x)) b(b(x)) (2)
c(c(x)) a(c(x)) (3)
a(c(x)) c(b(x)) (4)
b(a(x)) c(b(x)) (5)
b(b(x)) a(c(x)) (6)
c(c(x)) c(b(x)) (7)

The underlying signature is as follows:

{c/1, a/1, b/1}

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:

c(a(x)) b(c(x)) (1)
c(b(x)) b(b(x)) (2)
c(c(x)) a(c(x)) (3)
a(c(x)) c(b(x)) (4)
b(a(x)) c(b(x)) (5)
b(b(x)) a(c(x)) (6)
c(c(x)) c(b(x)) (7)
c(b(x)) a(c(x)) (8)
a(c(x)) b(b(x)) (9)
b(a(x)) b(b(x)) (10)
b(b(x)) c(b(x)) (11)
c(c(x)) b(b(x)) (12)

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 = a(c(a(f205)))
a(b(c(f205)))
= t1

t0 = a(c(a(f205)))
b(b(a(f205)))
b(c(b(f205)))
= t2

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