Certification Problem

Input (COPS 413)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

{f/3, g/1, p/2, h/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2021)

1 Redundant Rules Transformation

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

g(x) → h(x) (3)
f(x,y,z) → f(x,g(x),z) (2)
f(x,g(x),y) → p(h(x),y) (1)
f(x,y,z) → f(x,h(x),z) (4)
f(x,y,z) → p(h(x),z) (5)

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

1.1 Redundant Rules Transformation

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

f(x,y,z) → p(h(x),z) (5)
f(x,y,z) → f(x,h(x),z) (4)
f(x,y,z) → f(x,g(x),z) (2)
g(x) → h(x) (3)

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

1.1.1 Redundant Rules Transformation

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

f(x,y,z) → p(h(x),z) (5)
f(x,y,z) → f(x,h(x),z) (4)
f(x,y,z) → f(x,g(x),z) (2)
g(x) → h(x) (3)

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

1.1.1.1 Parallel Closed

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