Certification Problem

Input (COPS 19)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

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

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:

h(x,y) c (5)
f(x) h(x,x) (4)
a b (3)
g(b) c (2)
g(a) f(g(a)) (1)
f(x) c (6)
g(a) h(g(a),g(a)) (7)
g(a) f(g(b)) (8)
g(a) f(f(g(a))) (9)

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

1.1 Critical Pair Closing System

Confluence is proven using the following terminating critical-pair-closing-system R:

h(x,y) c (5)
g(b) c (2)
f(x) c (6)
f(x) h(x,x) (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c] = 0
[g(x1)] = 1 · x1 + 1
[b] = 3
[h(x1, x2)] = 4 · x1 + 2 · x2 + 0
[f(x1)] = 6 · x1 + 0
all of the following rules can be deleted.
g(b) c (2)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c] = 0
[h(x1, x2)] = 2 · x1 + 2 · x2 + 0
[f(x1)] = 6 · x1 + 2
all of the following rules can be deleted.
f(x) c (6)
f(x) h(x,x) (4)

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c] = 0
[h(x1, x2)] = 4 · x1 + 4 · x2 + 2
all of the following rules can be deleted.
h(x,y) c (5)

1.1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.