Certification Problem

Input (COPS 74)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

{a/0, c/0, b/0, f/2, d/0}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2022)

1 Redundant Rules Transformation

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

d f(c,b) (7)
d f(a,c) (6)
f(c,x) f(c,c) (5)
f(x,c) f(c,c) (4)
f(a,b) d (3)
b c (2)
a c (1)
d f(c,c) (8)
f(a,b) f(c,b) (9)
f(a,b) f(a,c) (10)

All redundant rules that were added or removed can be simulated in 2 steps .

1.1 Strongly closed

Confluence is proven since the TRS is strongly closed. The joins can be performed using 7 step(s).