Certification Problem
Input (TPDB SRS_Standard/Zantema_06/02)
The rewrite relation of the following TRS is considered.
|
a(p(x1)) |
→ |
p(a(A(x1))) |
(1) |
|
a(A(x1)) |
→ |
A(a(x1)) |
(2) |
|
p(A(A(x1))) |
→ |
a(p(x1)) |
(3) |
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
|
p(a(x1)) |
→ |
A(a(p(x1))) |
(4) |
|
A(a(x1)) |
→ |
a(A(x1)) |
(5) |
|
A(A(p(x1))) |
→ |
p(a(x1)) |
(6) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
p#(a(x1)) |
→ |
A#(a(p(x1))) |
(7) |
|
p#(a(x1)) |
→ |
p#(x1) |
(8) |
|
A#(a(x1)) |
→ |
A#(x1) |
(9) |
|
A#(A(p(x1))) |
→ |
p#(a(x1)) |
(10) |
1.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [p(x1)] |
= |
3 · x1
|
| [a(x1)] |
= |
1 + 1 · x1
|
| [A(x1)] |
= |
2 + 1 · x1
|
| [p#(x1)] |
= |
3 · x1
|
| [A#(x1)] |
= |
2 + 1 · x1
|
the
pairs
|
p#(a(x1)) |
→ |
p#(x1) |
(8) |
|
A#(a(x1)) |
→ |
A#(x1) |
(9) |
|
A#(A(p(x1))) |
→ |
p#(a(x1)) |
(10) |
and
the
rule
|
A(A(p(x1))) |
→ |
p(a(x1)) |
(6) |
could be deleted.
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 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [p#(x1)] |
= |
1 + 1 · x1
|
| [a(x1)] |
= |
1 + 1 · x1
|
| [A#(x1)] |
= |
1 |
| [p(x1)] |
= |
1 + 1 · x1
|
| [A(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
|
p#(a(x1)) |
→ |
A#(a(p(x1))) |
(7) |
could be deleted.
1.1.1.1.1.1 P is empty
There are no pairs anymore.