Certification Problem

Input (COPS 3)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

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

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2023)

1 Persistent Decomposition (Many-Sorted)

Confluence is proven, because the maximal systems induced by the sorts in the following many-sorted sort attachment are confluent.
f : 2 ⨯ 2 → 0
a : 2 → 2
b : 2 → 2
g : 2 ⨯ 1 → 1
c : 3 → 2
The subsystems are

(1.1)

f(a(x),x) → f(x,a(x)) (1)
f(b(x),x) → f(x,b(x)) (2)
a(x) → b(x) (5)

(1.2)

g(b(x),y) → g(a(a(x)),y) (3)
g(c(x),y) → y (4)
a(x) → b(x) (5)

1.1 Redundant Rules Transformation

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

a(x) → b(x) (5)
f(b(x),x) → f(x,b(x)) (2)
f(a(x),x) → f(x,a(x)) (1)
f(a(x),x) → f(x,b(x)) (6)

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

1.1.1 Locally confluent and terminating

Confluence is proven by showing local confluence and termination.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f(x1, x2)] = 4 · x1 + 3 · x2 + 2
[a(x1)] = 4 · x1 + 3
[b(x1)] = 4 · x1 + 2
all of the following rules can be deleted.
a(x) → b(x) (5)
f(b(x),x) → f(x,b(x)) (2)
f(a(x),x) → f(x,a(x)) (1)
f(a(x),x) → f(x,b(x)) (6)

1.1.1.1.1 R is empty

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

1.1.1.2 Local Confluence Proof

All critical pairs are joinable which can be seen by computing normal forms of all critical pairs.

1.2 Critical Pair Closing System

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

There are no rules.

1.2.1 R is empty

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