YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

a1 → b1
a1 → c1
b1 → b2
c1 → c2
a2 → b2
a2 → c2
b2 → b3
c2 → c3
a3 → b3
a3 → c3
b3 → b4
c3 → c4
a4 → b4
a4 → c4
b4 → b5
c4 → c5
a5 → b6
b5 → b6
c5 → b6

Proof

1 Critical Pair Closing System

Confluence is proven using the following terminating critical-pair-closing-system R:

b1 → b2
b2 → b3
b3 → b4
b4 → b5
b5 → b6
c1 → c2
c2 → c3
c3 → c4
c4 → c5
c5 → b6

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c2] = 4
[b6] = 0
[b1] = 0
[b2] = 0
[c4] = 0
[c3] = 0
[b5] = 0
[b4] = 0
[c5] = 0
[b3] = 0
[c1] = 4
the rules
b1 → b2
b2 → b3
b3 → b4
b4 → b5
b5 → b6
c1 → c2
c3 → c4
c4 → c5
c5 → b6
remain.

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[c4] = 4
[c3] = 4
[b5] = 0
[b4] = 0
[c5] = 0
[b3] = 0
[c1] = 0
the rules
b1 → b2
b2 → b3
b3 → b4
b4 → b5
b5 → b6
c1 → c2
c3 → c4
c5 → b6
remain.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[c4] = 0
[c3] = 0
[b5] = 0
[b4] = 0
[c5] = 1
[b3] = 0
[c1] = 0
the rules
b1 → b2
b2 → b3
b3 → b4
b4 → b5
b5 → b6
c1 → c2
c3 → c4
remain.

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[c4] = 0
[c3] = 1
[b5] = 0
[b4] = 0
[b3] = 0
[c1] = 0
the rules
b1 → b2
b2 → b3
b3 → b4
b4 → b5
b5 → b6
c1 → c2
remain.

1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[b5] = 0
[b4] = 0
[b3] = 0
[c1] = 1
the rules
b1 → b2
b2 → b3
b3 → b4
b4 → b5
b5 → b6
remain.

1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b6] = 0
[b1] = 4
[b2] = 4
[b5] = 0
[b4] = 0
[b3] = 4
the rules
b1 → b2
b2 → b3
b4 → b5
b5 → b6
remain.

1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b6] = 0
[b1] = 0
[b2] = 0
[b5] = 4
[b4] = 4
[b3] = 0
the rules
b1 → b2
b2 → b3
b4 → b5
remain.

1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b1] = 0
[b2] = 0
[b5] = 0
[b4] = 1
[b3] = 0
the rules
b1 → b2
b2 → b3
remain.

1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b1] = 4
[b2] = 4
[b3] = 0
the rule
b1 → b2
remains.

1.1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b1] = 1
[b2] = 0
all rules could be removed.

1.1.1.1.1.1.1.1.1.1.1.1 R is empty

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

Tool configuration

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