YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

W(W(x)) W(x)
B(I(x)) W(x)
W(B(x)) B(x)
F(H(x),y) F(H(x),G(y))
F(x,I(y)) F(G(x),I(y))
G(x) x

Proof

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)
B(I(x)) W(x)
W(B(x)) B(x)
G(x) x

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)
B(I(x)) W(x)

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
the rule
W(W(x)) W(x)
remains.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[W(x1)] = 2 · x1 + 5
all rules could be removed.

1.1.1.1.1 R is empty

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

Tool configuration

csi