Certification Problem

Input (COPS 99)

We consider the TRS containing the following rules:

W(W(x)) W(x) (1)
B(I(x)) W(x) (2)
W(B(x)) B(x) (3)
F(H(x),y) F(H(x),G(y)) (4)
F(x,I(y)) F(G(x),I(y)) (5)
G(x) x (6)

The underlying signature is as follows:

{W/1, B/1, I/1, F/2, H/1, G/1}

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:

W(W(x)) W(x) (1)
B(I(x)) W(x) (2)
W(B(x)) B(x) (3)
G(x) x (6)

All redundant rules that were added or removed can be simulated in 4 steps .

1.1 Critical Pair Closing System

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

W(W(x)) W(x) (1)
B(I(x)) W(x) (2)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[I(x1)] = 1 · x1 + 6
[W(x1)] = 1 · x1 + 0
[B(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
B(I(x)) W(x) (2)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[W(x1)] = 2 · x1 + 5
all of the following rules can be deleted.
W(W(x)) W(x) (1)

1.1.1.1.1 R is empty

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