Certification Problem

Input (COPS 423)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

{g/2, f/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,y) → h(g(f(x),f(y))) (4)
h(x) → g(f(x),x) (3)
f(x) → g(x,f(x)) (2)

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

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).