Certification Problem

Input (COPS 984)

We consider the TRS containing the following rules:

a(s(x)) → s(a(x)) (1)
b(a(b(s(x)))) → a(b(s(a(x)))) (2)
b(a(b(b(x)))) → a(b(a(b(x)))) (3)
a(b(a(a(x)))) → b(a(b(a(x)))) (4)

The underlying signature is as follows:

{a/1, s/1, b/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2020)

1 Critical Pair Closing System

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

a(s(x)) → s(a(x)) (1)
b(a(b(s(x)))) → a(b(s(a(x)))) (2)
a(b(a(a(x)))) → b(a(b(a(x)))) (4)
b(a(b(b(x)))) → a(b(a(b(x)))) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b(x1)] = 2 · x1 + 0
[a(x1)] = 2 · x1 + 0
[s(x1)] = 1 · x1 + 2
all of the following rules can be deleted.
a(s(x)) → s(a(x)) (1)
b(a(b(s(x)))) → a(b(s(a(x)))) (2)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 1 0
0 0 1
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[a(x1)] =
1 0 1
0 1 0
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
a(b(a(a(x)))) → b(a(b(a(x)))) (4)
b(a(b(b(x)))) → a(b(a(b(x)))) (3)

1.1.1.1 R is empty

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