Certification Problem

Input (COPS 412)

We consider the TRS containing the following rules:

f(x,y) f(g(x),g(x)) (1)
f(x,x) a (2)
g(x) x (3)

The underlying signature is as follows:

{f/2, g/1, 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:

g(x) x (3)
f(x,x) a (2)
f(x,y) f(g(x),g(x)) (1)
f(x,y) f(g(x),x) (4)
f(x,y) f(x,g(x)) (5)
f(x,y) a (6)
f(x,y) f(g(g(x)),g(g(x))) (7)

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) f(g(g(x)),g(g(x))) (7)
f(x,y) a (6)
f(x,y) f(x,g(x)) (5)
f(x,y) f(g(x),x) (4)
f(x,y) f(g(x),g(x)) (1)
g(x) 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) f(g(g(x)),g(g(x))) (7)
f(x,y) a (6)
f(x,y) f(x,g(x)) (5)
f(x,y) f(g(x),x) (4)
f(x,y) f(g(x),g(x)) (1)
g(x) x (3)
f(x,y) f(g(g(g(g(x)))),g(g(g(g(x))))) (8)
f(x,y) f(g(g(x)),g(g(g(x)))) (9)
f(x,y) f(g(g(g(x))),g(g(x))) (10)
f(x,y) f(g(g(g(x))),g(g(g(x)))) (11)
f(x,y) f(g(g(x)),g(x)) (12)
f(x,y) f(g(x),g(g(x))) (13)
f(x,y) f(x,x) (14)

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).