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 AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a(x1)] |
= |
1 · x1 + 39 |
| [b(x1)] |
= |
1 · x1 + 26 |
| [d(x1)] |
= |
1 · x1 + 11 |
| [c(x1)] |
= |
1 · x1 + 17 |
| [f(x1)] |
= |
1 · x1 + 39 |
all of the following rules can be deleted.
|
b(b(x1)) |
→ |
c(c(c(x1))) |
(4) |
|
c(c(x1)) |
→ |
d(d(d(x1))) |
(5) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(a(x1)) |
→ |
b#(b(b(x1))) |
(8) |
|
a#(a(x1)) |
→ |
b#(b(x1)) |
(9) |
|
a#(a(x1)) |
→ |
b#(x1) |
(10) |
|
a#(x1) |
→ |
c#(d(x1)) |
(11) |
|
b#(b(b(x1))) |
→ |
a#(f(x1)) |
(12) |
|
b#(b(b(x1))) |
→ |
f#(x1) |
(13) |
|
c#(d(d(x1))) |
→ |
f#(x1) |
(14) |
|
f#(f(x1)) |
→ |
f#(a(x1)) |
(15) |
|
f#(f(x1)) |
→ |
a#(x1) |
(16) |
1.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [a(x1)] |
= |
3 + 1 · x1
|
| [b(x1)] |
= |
2 + 1 · x1
|
| [d(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
3 + 1 · x1
|
| [f(x1)] |
= |
3 + 1 · x1
|
| [a#(x1)] |
= |
2 + 2 · x1
|
| [b#(x1)] |
= |
2 · x1
|
| [c#(x1)] |
= |
1 + 2 · x1
|
| [f#(x1)] |
= |
2 · x1
|
the
pairs
|
a#(a(x1)) |
→ |
b#(b(x1)) |
(9) |
|
a#(a(x1)) |
→ |
b#(x1) |
(10) |
|
a#(x1) |
→ |
c#(d(x1)) |
(11) |
|
b#(b(b(x1))) |
→ |
f#(x1) |
(13) |
|
c#(d(d(x1))) |
→ |
f#(x1) |
(14) |
|
f#(f(x1)) |
→ |
a#(x1) |
(16) |
and
no rules
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.