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 |
→ |
b7 |
| b7 |
→ |
b1 |
| b7 |
→ |
c1 |
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 |
| c1 |
→ |
c2 |
| c2 |
→ |
c3 |
| c3 |
→ |
c4 |
| c4 |
→ |
c5 |
| c5 |
→ |
c6 |
| c6 |
→ |
b7 |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b7] |
= |
0 |
| [c2] |
= |
4 |
| [b6] |
= |
0 |
| [b1] |
= |
0 |
| [b2] |
= |
0 |
| [c6] |
= |
0 |
| [c4] |
= |
1 |
| [c3] |
= |
4 |
| [b5] |
= |
0 |
| [b4] |
= |
0 |
| [c5] |
= |
0 |
| [b3] |
= |
0 |
| [c1] |
= |
4 |
the
rules
| b1 |
→ |
b2 |
| b2 |
→ |
b3 |
| b3 |
→ |
b4 |
| b4 |
→ |
b5 |
| b5 |
→ |
b6 |
| b6 |
→ |
b7 |
| c1 |
→ |
c2 |
| c2 |
→ |
c3 |
| c5 |
→ |
c6 |
| c6 |
→ |
b7 |
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 |
| [c3] |
= |
0 |
| [b5] |
= |
0 |
| [b4] |
= |
0 |
| [c5] |
= |
4 |
| [b3] |
= |
0 |
| [c1] |
= |
0 |
the
rules
| b1 |
→ |
b2 |
| b2 |
→ |
b3 |
| b3 |
→ |
b4 |
| b4 |
→ |
b5 |
| b5 |
→ |
b6 |
| b6 |
→ |
b7 |
| c1 |
→ |
c2 |
| c2 |
→ |
c3 |
| c5 |
→ |
c6 |
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 |
| [c3] |
= |
0 |
| [b5] |
= |
0 |
| [b4] |
= |
0 |
| [c5] |
= |
1 |
| [b3] |
= |
0 |
| [c1] |
= |
0 |
the
rules
| b1 |
→ |
b2 |
| b2 |
→ |
b3 |
| b3 |
→ |
b4 |
| b4 |
→ |
b5 |
| b5 |
→ |
b6 |
| b6 |
→ |
b7 |
| c1 |
→ |
c2 |
| c2 |
→ |
c3 |
remain.
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b7] |
= |
0 |
| [c2] |
= |
4 |
| [b6] |
= |
0 |
| [b1] |
= |
0 |
| [b2] |
= |
0 |
| [c3] |
= |
0 |
| [b5] |
= |
0 |
| [b4] |
= |
0 |
| [b3] |
= |
0 |
| [c1] |
= |
4 |
the
rules
| b1 |
→ |
b2 |
| b2 |
→ |
b3 |
| b3 |
→ |
b4 |
| b4 |
→ |
b5 |
| b5 |
→ |
b6 |
| b6 |
→ |
b7 |
| c1 |
→ |
c2 |
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 |
| [b5] |
= |
0 |
| [b4] |
= |
0 |
| [b3] |
= |
0 |
| [c1] |
= |
1 |
the
rules
| b1 |
→ |
b2 |
| b2 |
→ |
b3 |
| b3 |
→ |
b4 |
| b4 |
→ |
b5 |
| b5 |
→ |
b6 |
| b6 |
→ |
b7 |
remain.
1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b7] |
= |
0 |
| [b6] |
= |
0 |
| [b1] |
= |
4 |
| [b2] |
= |
4 |
| [b5] |
= |
4 |
| [b4] |
= |
4 |
| [b3] |
= |
4 |
the
rules
| b1 |
→ |
b2 |
| b2 |
→ |
b3 |
| b3 |
→ |
b4 |
| b4 |
→ |
b5 |
| b6 |
→ |
b7 |
remain.
1.1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b7] |
= |
0 |
| [b6] |
= |
1 |
| [b1] |
= |
0 |
| [b2] |
= |
0 |
| [b5] |
= |
0 |
| [b4] |
= |
0 |
| [b3] |
= |
0 |
the
rules
| b1 |
→ |
b2 |
| b2 |
→ |
b3 |
| b3 |
→ |
b4 |
| b4 |
→ |
b5 |
remain.
1.1.1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b1] |
= |
4 |
| [b2] |
= |
4 |
| [b5] |
= |
0 |
| [b4] |
= |
0 |
| [b3] |
= |
4 |
the
rules
remain.
1.1.1.1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b1] |
= |
0 |
| [b2] |
= |
0 |
| [b5] |
= |
0 |
| [b4] |
= |
1 |
| [b3] |
= |
0 |
the
rules
remain.
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 |
| [b3] |
= |
0 |
the
rule
remains.
1.1.1.1.1.1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
all rules could be removed.
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
- version: csi 1.2.5 [hg: unknown]
- strategy:
(cr -cpcs2 -cpcscert; ((( matrix -dim 1 -ib 3 -ob 5 | matrix -dim 2 -ib 2 -ob 3 | matrix -dim 3 -ib 1 -ob 1 | matrix -dim 3 -ib 1 -ob 3 | fail)[2]*);((dp;edg[0.5]?;(sccs | (sc || sct || {ur?;( (matrix -dp -ur -dim 1 -ib 3 -ob 5 | matrix -dp -ur -dim 2 -ib 2 -ob 3 | matrix -dp -ur -dim 3 -ib 1 -ob 1 | matrix -dp -ur -dim 3 -ib 1 -ob 3) || (kbo -ur -af | lpo -ur -af) || ( arctic -dp -ur -dim 2 -ib 2 -ob 2[2] | fail) || ( arctic -bz -dp -ur -dim 2 -ib 2 -ob 2[2] | fail) || fail) }restore || fail;(bounds -dp -rfc -qc || bounds -dp -all -rfc -qc || bounds -rfc -qc)[1] || fail ))*[6])! || (( kbo || (lpo | fail;(ref;lpo)) || fail;(bounds -rfc -qc) || fail)*[7])! || (rev;((dp;edg[0.5]?;(sccs | (sc || sct || {ur?;( (matrix -dp -ur -dim 1 -ib 3 -ob 5 | matrix -dp -ur -dim 2 -ib 2 -ob 3 | matrix -dp -ur -dim 3 -ib 1 -ob 1 | matrix -dp -ur -dim 3 -ib 1 -ob 3) || (kbo -ur -af | lpo -ur -af) || ( arctic -dp -ur -dim 2 -ib 2 -ob 2[2] | fail) || ( arctic -bz -dp -ur -dim 2 -ib 2 -ob 2[2] | fail) || fail) }restore || fail;(bounds -dp -rfc -qc || bounds -dp -all -rfc -qc || bounds -rfc -qc)[1] || fail ))*[6])! || (( kbo || (lpo | fail;(ref;lpo)) || fail;(bounds -rfc -qc) || fail)*[7])!)))))!