Certification Problem

Input (COPS 40)

We consider the TRS containing the following rules:

f(a) b (1)
f(a) f(c) (2)
a d (3)
f(d) b (4)
f(c) b (5)
d c (6)

The underlying signature is as follows:

{f/1, a/0, b/0, c/0, d/0}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2023)

1 Redundant Rules Transformation

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

d c (6)
f(c) b (5)
f(d) b (4)
a d (3)
f(a) f(c) (2)
f(a) b (1)
a c (7)

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:

f(c) b (5)
d c (6)
f(d) b (4)

1.1.1 Rule Removal

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

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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