Certification Problem

Input (COPS 16)

We consider the TRS containing the following rules:

f(x,x) g(x) (1)
f(x,g(x)) b (2)
h(c,y) f(h(y,c),h(y,y)) (3)

The underlying signature is as follows:

{f/2, g/1, b/0, h/2, c/0}

Property / Task

Prove or disprove confluence.

Answer / Result

No.

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:

f(x,x) g(x) (1)
f(x,g(x)) b (2)
h(c,y) f(h(y,c),h(y,y)) (3)
h(c,c) g(h(c,c)) (4)
f(x,f(x,x)) b (5)

All redundant rules that were added or removed can be simulated in 3 steps .

1.1 Non-Joinable Fork

The system is not confluent due to the following forking derivations.

t0 = h(c,c)
f(h(c,c),h(c,c))
f(h(c,c),g(h(c,c)))
b
= t3

t0 = h(c,c)
g(h(c,c))
g(g(h(c,c)))
g(g(g(h(c,c))))
= t3

The two resulting terms cannot be joined for the following reason: