Certification Problem

Input (COPS 65)

We consider the TRS containing the following rules:

h(f,a,a) → h(g,a,a) (1)
h(g,a,a) → h(f,a,a) (2)
a → a' (3)
h(x,a',y) → h(x,y,y) (4)
g → f (5)
f → g (6)

The underlying signature is as follows:

{h/3, f/0, a/0, g/0, a'/0}

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:

f → g (6)
g → f (5)
h(x,a',y) → h(x,y,y) (4)
a → a' (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:

f → g (6)
g → f (5)
h(x,a',y) → h(x,y,y) (4)
a → a' (3)
f → f (7)
g → g (8)

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

1.1.1 Development Closed

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