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