YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

a(x) x
a(b(x)) c(b(b(a(a(x)))))
b(x) c(x)
c(c(x)) x

Proof

1 Redundant Rules Transformation

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

c(c(x)) x
b(x) c(x)
a(b(x)) c(b(b(a(a(x)))))
a(x) x
a(b(x)) c(b(c(a(a(x)))))
a(b(x)) c(c(b(a(a(x)))))
a(b(x)) c(b(b(a(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:

a(x) x
b(x) c(x)
c(c(x)) x

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b(x1)] = 1 · x1 + 0
[a(x1)] = 1 · x1 + 1
[c(x1)] = 1 · x1 + 0
the rules
b(x) c(x)
c(c(x)) x
remain.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b(x1)] = 4 · x1 + 4
[c(x1)] = 2 · x1 + 4
the rule
b(x) c(x)
remains.

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b(x1)] = 4 · x1 + 1
[c(x1)] = 4 · x1 + 0
all rules could be removed.

1.1.1.1.1.1 R is empty

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

Tool configuration

csi