YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

F(G(x,A,B)) x
G(F(H(C,D)),x,y) H(K1(x),K2(y))
K1(A) C
K2(B) D

Proof

1 Redundant Rules Transformation

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

K2(B) D
K1(A) C
G(F(H(C,D)),x,y) H(K1(x),K2(y))
F(G(x,A,B)) 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:

K1(A) C
K2(B) D

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[C] = 0
[K2(x1)] = 4 · x1 + 1
[B] = 0
[D] = 0
[A] = 0
[K1(x1)] = 2 · x1 + 0
the rule
K1(A) C
remains.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[C] = 0
[A] = 4
[K1(x1)] = 1 · x1 + 1
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