Certification Problem

Input (COPS 948)

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)))) c(s(x)) (3)
c(s(x)) a(b(a(b(x)))) (4)
a(b(a(a(x)))) b(a(b(a(x)))) (5)

The underlying signature is as follows:

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

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

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:

a(b(a(a(x)))) b(a(b(a(x)))) (5)
c(s(x)) a(b(a(b(x)))) (4)
b(a(b(b(x)))) c(s(x)) (3)
b(a(b(s(x)))) a(b(s(a(x)))) (2)
a(s(x)) s(a(x)) (1)
b(a(b(b(x)))) a(b(a(b(x)))) (6)

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:

a(b(a(a(x)))) b(a(b(a(x)))) (5)
b(a(b(b(x)))) a(b(a(b(x)))) (6)
c(s(x)) a(b(a(b(x)))) (4)
b(a(b(s(x)))) a(b(s(a(x)))) (2)
a(s(x)) s(a(x)) (1)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a(x1)] = 1 · x1 + 0
[c(x1)] = 5 · x1 + 1
[s(x1)] = 2 · x1 + 0
[b(x1)] = 1 · x1 + 0
all of the following rules can be deleted.
c(s(x)) a(b(a(b(x)))) (4)

1.1.1.1 Rule Removal

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

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 1 0
1 0 0
1 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[b(x1)] =
1 0 1
1 0 0
1 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
a(b(a(a(x)))) b(a(b(a(x)))) (5)
b(a(b(b(x)))) a(b(a(b(x)))) (6)

1.1.1.1.1.1 R is empty

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