Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z123)
The rewrite relation of the following TRS is considered.
|
a(a(x1)) |
→ |
b(b(b(x1))) |
(1) |
|
a(x1) |
→ |
d(c(d(x1))) |
(2) |
|
b(b(b(x1))) |
→ |
a(f(x1)) |
(3) |
|
b(b(x1)) |
→ |
c(c(c(x1))) |
(4) |
|
c(c(x1)) |
→ |
d(d(d(x1))) |
(5) |
|
c(d(d(x1))) |
→ |
f(x1) |
(6) |
|
f(f(x1)) |
→ |
f(a(x1)) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1)] |
= |
1 · x1 + 21 |
| [b(x1)] |
= |
1 · x1 + 14 |
| [c(x1)] |
= |
1 · x1 + 9 |
| [a(x1)] |
= |
1 · x1 + 21 |
| [d(x1)] |
= |
1 · x1 + 6 |
all of the following rules can be deleted.
|
b(b(x1)) |
→ |
c(c(c(x1))) |
(4) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [f(x1)] |
= |
6 · x1 +
-∞ |
| [b(x1)] |
= |
4 · x1 +
-∞ |
| [c(x1)] |
= |
6 · x1 +
-∞ |
| [a(x1)] |
= |
6 · x1 +
-∞ |
| [d(x1)] |
= |
0 · x1 +
-∞ |
all of the following rules can be deleted.
|
c(c(x1)) |
→ |
d(d(d(x1))) |
(5) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1)] |
= |
8 · x1 + 21 |
| [b(x1)] |
= |
4 · x1 + 9 |
| [c(x1)] |
= |
2 · x1 + 0 |
| [a(x1)] |
= |
8 · x1 + 21 |
| [d(x1)] |
= |
2 · x1 + 4 |
all of the following rules can be deleted.
|
a(x1) |
→ |
d(c(d(x1))) |
(2) |
|
c(d(d(x1))) |
→ |
f(x1) |
(6) |
1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(a(x1)) |
→ |
b#(x1) |
(8) |
|
a#(a(x1)) |
→ |
b#(b(x1)) |
(9) |
|
a#(a(x1)) |
→ |
b#(b(b(x1))) |
(10) |
|
b#(b(b(x1))) |
→ |
f#(x1) |
(11) |
|
b#(b(b(x1))) |
→ |
a#(f(x1)) |
(12) |
|
f#(f(x1)) |
→ |
a#(x1) |
(13) |
|
f#(f(x1)) |
→ |
f#(a(x1)) |
(14) |
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.