Certification Problem

Input (COPS 87)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

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

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

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:

g(x) → p(p(h(x))) (4)
f(x) → h(f(x)) (3)
h(x) → p(h(x)) (2)
f(x) → g(f(x)) (1)
g(x) → p(p(p(h(x)))) (5)
f(x) → h(h(f(x))) (6)
f(x) → p(h(f(x))) (7)
f(x) → h(g(f(x))) (8)
h(x) → p(p(h(x))) (9)
f(x) → p(p(h(f(x)))) (10)
f(x) → g(h(f(x))) (11)
f(x) → g(g(f(x))) (12)

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

1.1 Development Closed

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