Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z095)
The rewrite relation of the following TRS is considered.
|
a(a(b(b(x1)))) |
→ |
b(b(b(a(a(a(a(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
|
b(b(a(a(x1)))) |
→ |
a(a(a(a(a(b(b(b(x1)))))))) |
(2) |
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 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(b(a(a(x1)))) |
→ |
b#(b(b(x1))) |
(3) |
|
b#(b(a(a(x1)))) |
→ |
b#(b(x1)) |
(4) |
|
b#(b(a(a(x1)))) |
→ |
b#(x1) |
(5) |
1.1.1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
| [b#(x1)] |
= |
+
|
|
-∞ |
-∞ |
0 |
|
-∞ |
-∞ |
-∞ |
|
-∞ |
-∞ |
-∞ |
|
|
· x1
|
| [b(x1)] |
= |
+ · x1
|
| [a(x1)] |
= |
+ · x1
|
the
pairs
|
b#(b(a(a(x1)))) |
→ |
b#(b(b(x1))) |
(3) |
|
b#(b(a(a(x1)))) |
→ |
b#(x1) |
(5) |
could be deleted.
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):
| [a(x1)] |
= |
0 |
| [b(x1)] |
= |
1 + 1x1
|
| [b#(x1)] |
= |
0 |
We obtain the set of labeled pairs
|
b#1(b0(a0(a0(x1)))) |
→ |
b#1(b0(x1)) |
(6) |
|
b#1(b0(a0(a1(x1)))) |
→ |
b#0(b1(x1)) |
(7) |
and the set of labeled rules:
|
b1(b0(a0(a0(x1)))) |
→ |
a0(a0(a0(a0(a1(b0(b1(b0(x1)))))))) |
(8) |
|
b1(b0(a0(a1(x1)))) |
→ |
a0(a0(a0(a0(a0(b1(b0(b1(x1)))))))) |
(9) |
Innermost rewriting w.r.t. the following left-hand sides is considered:
|
b1(b0(a0(a0(x0)))) |
|
b1(b0(a0(a1(x0)))) |
1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.