Certification Problem
Input (TPDB SRS_Standard/Secret_06_SRS/multum1)
The rewrite relation of the following TRS is considered.
|
a(a(b(x1))) |
→ |
b(a(a(a(x1)))) |
(1) |
|
b(a(b(a(x1)))) |
→ |
a(b(b(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(x1)))) |
(3) |
|
a(b(a(b(x1)))) |
→ |
b(b(a(x1))) |
(4) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(a(a(x1))) |
→ |
a#(a(a(b(x1)))) |
(5) |
|
b#(a(a(x1))) |
→ |
a#(a(b(x1))) |
(6) |
|
b#(a(a(x1))) |
→ |
a#(b(x1)) |
(7) |
|
b#(a(a(x1))) |
→ |
b#(x1) |
(8) |
|
a#(b(a(b(x1)))) |
→ |
b#(b(a(x1))) |
(9) |
|
a#(b(a(b(x1)))) |
→ |
b#(a(x1)) |
(10) |
|
a#(b(a(b(x1)))) |
→ |
a#(x1) |
(11) |
1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [b#(x1)] |
= |
1 + 1 · x1
|
| [a(x1)] |
= |
1 · x1
|
| [a#(x1)] |
= |
1 · x1
|
| [b(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
a#(b(a(b(x1)))) |
→ |
b#(a(x1)) |
(10) |
|
a#(b(a(b(x1)))) |
→ |
a#(x1) |
(11) |
could be deleted.
1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the rationals with delta = 1/32
| [b#(x1)] |
= |
1/2 + 3/4 · x1
|
| [a(x1)] |
= |
1/4 + 1 · x1
|
| [a#(x1)] |
= |
1/2 + 1/2 · x1
|
| [b(x1)] |
= |
1/4 + 3/2 · x1
|
the
pairs
|
b#(a(a(x1))) |
→ |
a#(a(b(x1))) |
(6) |
|
b#(a(a(x1))) |
→ |
a#(b(x1)) |
(7) |
|
b#(a(a(x1))) |
→ |
b#(x1) |
(8) |
|
a#(b(a(b(x1)))) |
→ |
b#(b(a(x1))) |
(9) |
could be deleted.
1.1.1.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [b(x1)] |
= |
1 · x1
|
| [a(x1)] |
= |
1 · x1
|
| [b#(x1)] |
= |
1 + 2 · x1
|
| [a#(x1)] |
= |
1 · x1
|
the
pair
|
b#(a(a(x1))) |
→ |
a#(a(a(b(x1)))) |
(5) |
and
no rules
could be deleted.
1.1.1.1.1.1 P is empty
There are no pairs anymore.