Certification Problem
Input (TPDB SRS_Standard/Trafo_06/dup16)
The rewrite relation of the following TRS is considered.
|
a(a(a(a(x1)))) |
→ |
b(b(b(b(b(b(x1)))))) |
(1) |
|
b(b(b(b(x1)))) |
→ |
c(c(c(c(c(c(x1)))))) |
(2) |
|
c(c(c(c(x1)))) |
→ |
d(d(d(d(d(d(x1)))))) |
(3) |
|
b(b(x1)) |
→ |
d(d(d(d(x1)))) |
(4) |
|
c(c(d(d(d(d(x1)))))) |
→ |
a(a(x1)) |
(5) |
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(a(x1)))) |
→ |
b(b(b(b(b(b(x1)))))) |
(1) |
|
b(b(b(b(x1)))) |
→ |
c(c(c(c(c(c(x1)))))) |
(2) |
|
c(c(c(c(x1)))) |
→ |
d(d(d(d(d(d(x1)))))) |
(3) |
|
b(b(x1)) |
→ |
d(d(d(d(x1)))) |
(4) |
|
d(d(d(d(c(c(x1)))))) |
→ |
a(a(x1)) |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a(x1)] |
= |
1 · x1 + 66 |
| [b(x1)] |
= |
1 · x1 + 44 |
| [c(x1)] |
= |
1 · x1 + 29 |
| [d(x1)] |
= |
1 · x1 + 19 |
all of the following rules can be deleted.
|
b(b(b(b(x1)))) |
→ |
c(c(c(c(c(c(x1)))))) |
(2) |
|
c(c(c(c(x1)))) |
→ |
d(d(d(d(d(d(x1)))))) |
(3) |
|
b(b(x1)) |
→ |
d(d(d(d(x1)))) |
(4) |
|
d(d(d(d(c(c(x1)))))) |
→ |
a(a(x1)) |
(6) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a(x1)] |
= |
1 · x1 + 1 |
| [b(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
a(a(a(a(x1)))) |
→ |
b(b(b(b(b(b(x1)))))) |
(1) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.