Certification Problem

Input (COPS 121)

We consider the TRS containing the following rules:

f(g(x),g(y)) f(g(x),h(y)) (1)
f(h(x),g(y)) f(g(x),g(y)) (2)
f(g(x),h(y)) f(x,y) (3)
f(h(x),h(y)) f(y,x) (4)
f(x,y) f(y,x) (5)
g(x) h(x) (6)
h(x) g(x) (7)

The underlying signature is as follows:

{f/2, g/1, h/1}

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:

h(x) g(x) (7)
g(x) h(x) (6)
f(x,y) f(y,x) (5)
f(h(x),h(y)) f(y,x) (4)
f(g(x),h(y)) f(x,y) (3)

All redundant rules that were added or removed can be simulated in 1 steps .

1.1 Redundant Rules Transformation

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

h(x) g(x) (7)
g(x) h(x) (6)
f(x,y) f(y,x) (5)
f(h(x),h(y)) f(y,x) (4)
f(g(x),h(y)) f(x,y) (3)
f(h(y),g(x)) f(x,y) (8)
f(g(x),g(y)) f(x,y) (9)
f(h(x),g(y)) f(y,x) (10)
f(h(x85),g(x84)) f(x84,x85) (11)

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

1.1.1 Strongly closed

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