Certification Problem

Input (COPS 86)

We consider the TRS containing the following rules:

f(x) → g(x) (1)
f(x) → h(f(x)) (2)
h(f(x)) → h(g(x)) (3)
g(x) → h(g(x)) (4)

The underlying signature is as follows:

{f/1, 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:

g(x) → h(g(x)) (4)
f(x) → h(f(x)) (2)
f(x) → g(x) (1)

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:

g(x) → h(g(x)) (4)
f(x) → h(f(x)) (2)
f(x) → g(x) (1)
g(x) → h(h(g(x))) (5)
f(x) → h(h(f(x))) (6)
f(x) → h(g(x)) (7)

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

1.1.1 Parallel Closed

Confluence is proven since the TRS is (almost) parallel closed. The joins can be performed using 1 parallel step(s).