Certification Problem
Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-536)
The rewrite relation of the following TRS is considered.
a(a(b(x1))) |
→ |
c(b(a(a(a(x1))))) |
(1) |
a(c(x1)) |
→ |
b(a(x1)) |
(2) |
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(a(a(x1))) |
→ |
a(a(a(b(c(x1))))) |
(3) |
c(a(x1)) |
→ |
a(b(x1)) |
(4) |
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#(a(a(x1))) |
→ |
b#(c(x1)) |
(5) |
b#(a(a(x1))) |
→ |
c#(x1) |
(6) |
c#(a(x1)) |
→ |
b#(x1) |
(7) |
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
|
[a(x1)] |
= |
+ · x1
|
[c(x1)] |
= |
+ · x1
|
[c#(x1)] |
= |
+ · x1
|
[b(x1)] |
= |
+ · x1
|
the
pair
could be deleted.
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.