YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

0(1(2(3(4(x))))) 0(2(1(3(4(x)))))
0(5(1(2(4(3(x)))))) 0(5(2(1(4(3(x))))))
0(5(2(4(1(3(x)))))) 0(1(5(2(4(3(x))))))
0(5(3(1(2(4(x)))))) 0(1(5(3(2(4(x))))))
0(5(4(1(3(2(x)))))) 0(5(4(3(1(2(x))))))

Proof

1 Redundant Rules Transformation

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

0(5(4(1(3(2(x)))))) 0(5(4(3(1(2(x))))))
0(5(3(1(2(4(x)))))) 0(1(5(3(2(4(x))))))
0(5(2(4(1(3(x)))))) 0(1(5(2(4(3(x))))))
0(5(1(2(4(3(x)))))) 0(5(2(1(4(3(x))))))
0(1(2(3(4(x))))) 0(2(1(3(4(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:

There are no rules.

1.1.1 R is empty

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

Tool configuration

csi