Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z030)
The rewrite relation of the following TRS is considered.
|
b(a(b(a(a(a(x1)))))) |
→ |
a(a(a(b(a(b(a(b(a(x1))))))))) |
(1) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(a(a(b(a(b(x1)))))) |
→ |
a(b(a(b(a(b(a(a(a(x1))))))))) |
(2) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
aa(aa(ab(ba(ab(ba(x1)))))) |
→ |
ab(ba(ab(ba(ab(ba(aa(aa(aa(x1))))))))) |
(3) |
|
aa(aa(ab(ba(ab(bb(x1)))))) |
→ |
ab(ba(ab(ba(ab(ba(aa(aa(ab(x1))))))))) |
(4) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [aa(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
aa(aa(ab(ba(ab(bb(x1)))))) |
→ |
ab(ba(ab(ba(ab(ba(aa(aa(ab(x1))))))))) |
(4) |
1.1.1.1 Switch to Innermost Termination
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
aa#(aa(ab(ba(ab(ba(x1)))))) |
→ |
aa#(aa(aa(x1))) |
(5) |
|
aa#(aa(ab(ba(ab(ba(x1)))))) |
→ |
aa#(aa(x1)) |
(6) |
|
aa#(aa(ab(ba(ab(ba(x1)))))) |
→ |
aa#(x1) |
(7) |
1.1.1.1.1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the integers
| [aa#(x1)] |
= |
+
|
| -1 |
-∞ |
-1 |
|
-∞ |
-∞ |
-∞ |
|
-∞ |
-∞ |
-∞ |
|
|
· x1
|
| [aa(x1)] |
= |
+ · x1
|
| [ab(x1)] |
= |
+
|
|
-∞ |
-1 |
-1 |
| -1 |
-1 |
-∞ |
|
-∞ |
-1 |
-∞ |
|
|
· x1
|
| [ba(x1)] |
= |
+ · x1
|
the
pairs
|
aa#(aa(ab(ba(ab(ba(x1)))))) |
→ |
aa#(aa(aa(x1))) |
(5) |
|
aa#(aa(ab(ba(ab(ba(x1)))))) |
→ |
aa#(x1) |
(7) |
could be deleted.
1.1.1.1.1.1.1 Semantic Labeling Processor
The following interpretations form a
model
of the rules.
As carrier we take the set
{0,1}.
Symbols are labeled by the interpretation of their arguments using the interpretations
(modulo 2):
| [ba(x1)] |
= |
0 |
| [aa#(x1)] |
= |
0 |
| [ab(x1)] |
= |
0 |
| [aa(x1)] |
= |
1 + 1x1
|
We obtain the set of labeled pairs
|
aa#1(aa0(ab0(ba0(ab0(ba0(x1)))))) |
→ |
aa#1(aa0(x1)) |
(8) |
|
aa#1(aa0(ab0(ba0(ab0(ba1(x1)))))) |
→ |
aa#0(aa1(x1)) |
(9) |
and the set of labeled rules:
|
aa1(aa0(ab0(ba0(ab0(ba0(x1)))))) |
→ |
ab0(ba0(ab0(ba0(ab0(ba1(aa0(aa1(aa0(x1))))))))) |
(10) |
|
aa1(aa0(ab0(ba0(ab0(ba1(x1)))))) |
→ |
ab0(ba0(ab0(ba0(ab0(ba0(aa1(aa0(aa1(x1))))))))) |
(11) |
Innermost rewriting w.r.t. the following left-hand sides is considered:
|
aa1(aa0(ab0(ba0(ab0(ba0(x0)))))) |
|
aa1(aa0(ab0(ba0(ab0(ba1(x0)))))) |
1.1.1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.