Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z088)
The rewrite relation of the following TRS is considered.
|
a(a(x1)) |
→ |
a(b(b(b(x1)))) |
(1) |
|
b(a(x1)) |
→ |
b(b(c(x1))) |
(2) |
|
a(b(b(c(x1)))) |
→ |
a(a(a(b(x1)))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
aa(aa(x1)) |
→ |
ab(bb(bb(ba(x1)))) |
(4) |
|
aa(ab(x1)) |
→ |
ab(bb(bb(bb(x1)))) |
(5) |
|
aa(ac(x1)) |
→ |
ab(bb(bb(bc(x1)))) |
(6) |
|
ba(aa(x1)) |
→ |
bb(bc(ca(x1))) |
(7) |
|
ba(ab(x1)) |
→ |
bb(bc(cb(x1))) |
(8) |
|
ba(ac(x1)) |
→ |
bb(bc(cc(x1))) |
(9) |
|
ab(bb(bc(ca(x1)))) |
→ |
aa(aa(ab(ba(x1)))) |
(10) |
|
ab(bb(bc(cb(x1)))) |
→ |
aa(aa(ab(bb(x1)))) |
(11) |
|
ab(bb(bc(cc(x1)))) |
→ |
aa(aa(ab(bc(x1)))) |
(12) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [aa(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [ac(x1)] |
= |
1 · x1 + 2 |
| [bc(x1)] |
= |
1 · x1
|
| [ca(x1)] |
= |
1 · x1
|
| [cb(x1)] |
= |
1 · x1
|
| [cc(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
aa(ac(x1)) |
→ |
ab(bb(bb(bc(x1)))) |
(6) |
|
ba(ac(x1)) |
→ |
bb(bc(cc(x1))) |
(9) |
|
ab(bb(bc(cc(x1)))) |
→ |
aa(aa(ab(bc(x1)))) |
(12) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
aa#(aa(x1)) |
→ |
ab#(bb(bb(ba(x1)))) |
(13) |
|
aa#(aa(x1)) |
→ |
ba#(x1) |
(14) |
|
aa#(ab(x1)) |
→ |
ab#(bb(bb(bb(x1)))) |
(15) |
|
ab#(bb(bc(ca(x1)))) |
→ |
aa#(aa(ab(ba(x1)))) |
(16) |
|
ab#(bb(bc(ca(x1)))) |
→ |
aa#(ab(ba(x1))) |
(17) |
|
ab#(bb(bc(ca(x1)))) |
→ |
ab#(ba(x1)) |
(18) |
|
ab#(bb(bc(ca(x1)))) |
→ |
ba#(x1) |
(19) |
|
ab#(bb(bc(cb(x1)))) |
→ |
aa#(aa(ab(bb(x1)))) |
(20) |
|
ab#(bb(bc(cb(x1)))) |
→ |
aa#(ab(bb(x1))) |
(21) |
|
ab#(bb(bc(cb(x1)))) |
→ |
ab#(bb(x1)) |
(22) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
1st
component contains the
pair
|
ab#(bb(bc(cb(x1)))) |
→ |
ab#(bb(x1)) |
(22) |
|
ab#(bb(bc(ca(x1)))) |
→ |
ab#(ba(x1)) |
(18) |
1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [ba(x1)] |
= |
1 · x1
|
| [aa(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
| [bc(x1)] |
= |
1 · x1
|
| [ca(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [cb(x1)] |
= |
1 · x1
|
| [ab#(x1)] |
= |
1 · x1
|
together with the usable
rules
|
ba(aa(x1)) |
→ |
bb(bc(ca(x1))) |
(7) |
|
ba(ab(x1)) |
→ |
bb(bc(cb(x1))) |
(8) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.1.1.1.1 Switch to Innermost Termination
The TRS does not have overlaps with the pairs and is locally confluent:
20
Hence, it suffices to show innermost termination in the following.
1.1.1.1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [ab#(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 + 1 · x1
|
| [bc(x1)] |
= |
1 · x1
|
| [cb(x1)] |
= |
1 · x1
|
| [ca(x1)] |
= |
1 + 1 · x1
|
| [ba(x1)] |
= |
1 + 1 · x1
|
| [aa(x1)] |
= |
1 + 1 · x1
|
| [ab(x1)] |
= |
1 + 1 · x1
|
the
pair
|
ab#(bb(bc(ca(x1)))) |
→ |
ab#(ba(x1)) |
(18) |
could be deleted.
1.1.1.1.1.1.1.1 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
There are no rules.
1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
There are no lhss.
1.1.1.1.1.1.1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [ab#(x1)] |
= |
2 · x1
|
| [bb(x1)] |
= |
3 · x1
|
| [bc(x1)] |
= |
1 + 4 · x1
|
| [cb(x1)] |
= |
1 · x1
|
the
pair
|
ab#(bb(bc(cb(x1)))) |
→ |
ab#(bb(x1)) |
(22) |
could be deleted.
1.1.1.1.1.1.1.1.1.1.1 P is empty
There are no pairs anymore.