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 b5
a5 c5
b5 b6
c5 c6
a6 b6
a6 c6
b6 b7
c6 c7
a7 b7
a7 c7
b7 b8
c7 c8
a8 b8
a8 c8
b8 b9
c8 c9
a9 b9
a9 c9
b9 b10
c9 c10
a10 b11
b10 b11
c10 b11

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
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
c1 c2
c2 c3
c3 c4
c4 c5
c5 c6
c6 c7
c7 c8
c8 c9
c9 c10
c10 b11

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[c2] = 4
[b6] = 0
[b1] = 0
[b2] = 0
[c6] = 0
[b8] = 0
[c7] = 0
[b10] = 0
[c8] = 0
[c10] = 0
[c4] = 2
[c3] = 2
[c9] = 0
[b11] = 0
[b5] = 0
[b9] = 0
[b4] = 0
[c5] = 2
[b3] = 0
[c1] = 4
the rules
b1 b2
b2 b3
b3 b4
b4 b5
b5 b6
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
c1 c2
c3 c4
c4 c5
c6 c7
c7 c8
c8 c9
c9 c10
c10 b11
remain.

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[c6] = 4
[b8] = 0
[c7] = 4
[b10] = 0
[c8] = 4
[c10] = 0
[c4] = 0
[c3] = 0
[c9] = 4
[b11] = 0
[b5] = 0
[b9] = 0
[b4] = 0
[c5] = 0
[b3] = 0
[c1] = 0
the rules
b1 b2
b2 b3
b3 b4
b4 b5
b5 b6
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
c1 c2
c3 c4
c4 c5
c6 c7
c7 c8
c8 c9
c10 b11
remain.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[c6] = 0
[b8] = 0
[c7] = 0
[b10] = 0
[c8] = 0
[c10] = 1
[c4] = 0
[c3] = 0
[c9] = 0
[b11] = 0
[b5] = 0
[b9] = 0
[b4] = 0
[c5] = 0
[b3] = 0
[c1] = 0
the rules
b1 b2
b2 b3
b3 b4
b4 b5
b5 b6
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
c1 c2
c3 c4
c4 c5
c6 c7
c7 c8
c8 c9
remain.

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[c6] = 4
[b8] = 0
[c7] = 4
[b10] = 0
[c8] = 0
[c4] = 0
[c3] = 0
[c9] = 0
[b11] = 0
[b5] = 0
[b9] = 0
[b4] = 0
[c5] = 0
[b3] = 0
[c1] = 0
the rules
b1 b2
b2 b3
b3 b4
b4 b5
b5 b6
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
c1 c2
c3 c4
c4 c5
c6 c7
c8 c9
remain.

1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[c6] = 0
[b8] = 0
[c7] = 0
[b10] = 0
[c8] = 1
[c4] = 0
[c3] = 0
[c9] = 0
[b11] = 0
[b5] = 0
[b9] = 0
[b4] = 0
[c5] = 0
[b3] = 0
[c1] = 0
the rules
b1 b2
b2 b3
b3 b4
b4 b5
b5 b6
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
c1 c2
c3 c4
c4 c5
c6 c7
remain.

1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[c6] = 1
[b8] = 0
[c7] = 0
[b10] = 0
[c4] = 0
[c3] = 0
[b11] = 0
[b5] = 0
[b9] = 0
[b4] = 0
[c5] = 0
[b3] = 0
[c1] = 0
the rules
b1 b2
b2 b3
b3 b4
b4 b5
b5 b6
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
c1 c2
c3 c4
c4 c5
remain.

1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[b8] = 0
[b10] = 0
[c4] = 4
[c3] = 4
[b11] = 0
[b5] = 0
[b9] = 0
[b4] = 0
[c5] = 0
[b3] = 0
[c1] = 0
the rules
b1 b2
b2 b3
b3 b4
b4 b5
b5 b6
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
c1 c2
c3 c4
remain.

1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[b8] = 0
[b10] = 0
[c4] = 0
[c3] = 1
[b11] = 0
[b5] = 0
[b9] = 0
[b4] = 0
[b3] = 0
[c1] = 0
the rules
b1 b2
b2 b3
b3 b4
b4 b5
b5 b6
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
c1 c2
remain.

1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[c2] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[b8] = 0
[b10] = 0
[b11] = 0
[b5] = 0
[b9] = 0
[b4] = 0
[b3] = 0
[c1] = 1
the rules
b1 b2
b2 b3
b3 b4
b4 b5
b5 b6
b6 b7
b7 b8
b8 b9
b9 b10
b10 b11
remain.

1.1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[b6] = 1
[b1] = 4
[b2] = 4
[b8] = 0
[b10] = 0
[b11] = 0
[b5] = 1
[b9] = 0
[b4] = 4
[b3] = 4
the rules
b1 b2
b2 b3
b3 b4
b5 b6
b7 b8
b8 b9
b9 b10
b10 b11
remain.

1.1.1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 4
[b6] = 0
[b1] = 0
[b2] = 0
[b8] = 4
[b10] = 0
[b11] = 0
[b5] = 0
[b9] = 1
[b4] = 0
[b3] = 0
the rules
b1 b2
b2 b3
b3 b4
b5 b6
b7 b8
b10 b11
remain.

1.1.1.1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 0
[b6] = 0
[b1] = 0
[b2] = 0
[b8] = 0
[b10] = 1
[b11] = 0
[b5] = 0
[b4] = 0
[b3] = 0
the rules
b1 b2
b2 b3
b3 b4
b5 b6
b7 b8
remain.

1.1.1.1.1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b7] = 1
[b6] = 0
[b1] = 0
[b2] = 0
[b8] = 0
[b5] = 0
[b4] = 0
[b3] = 0
the rules
b1 b2
b2 b3
b3 b4
b5 b6
remain.

1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b1] = 0
[b2] = 0
[b4] = 0
[b3] = 1
the rule
b1 b2
remains.

1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 R is empty

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

Tool configuration

csi