YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

F(H(x),y) → G(H(x))
H(I(x)) → I(x)
F(I(x),y) → G(I(x))

Proof

1 Redundant Rules Transformation

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

F(I(x),y) → G(I(x))
H(I(x)) → I(x)
F(H(x),y) → G(H(x))

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(I(x),y) → G(I(x))
H(I(x)) → I(x)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[F(x1, x2)] = 5 · x1 + 4 · x2 + 6
[I(x1)] = 1 · x1 + 2
[H(x1)] = 1 · x1 + 1
[G(x1)] = 5 · x1 + 2
all rules could be removed.

1.1.1.1 R is empty

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

Tool configuration

csi