Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z071)
The rewrite relation of the following TRS is considered.
|
R(x1) |
→ |
r(x1) |
(1) |
|
r(p(x1)) |
→ |
p(p(r(P(x1)))) |
(2) |
|
r(r(x1)) |
→ |
x1 |
(3) |
|
r(P(P(x1))) |
→ |
P(P(r(x1))) |
(4) |
|
p(P(x1)) |
→ |
x1 |
(5) |
|
P(p(x1)) |
→ |
x1 |
(6) |
|
r(R(x1)) |
→ |
x1 |
(7) |
|
R(r(x1)) |
→ |
x1 |
(8) |
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
| [R(x1)] |
= |
1 · x1 + 1 |
| [r(x1)] |
= |
1 · x1
|
| [p(x1)] |
= |
1 · x1
|
| [P(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
R(x1) |
→ |
r(x1) |
(1) |
|
r(R(x1)) |
→ |
x1 |
(7) |
|
R(r(x1)) |
→ |
x1 |
(8) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [r(x1)] |
= |
1 · x1 + 1 |
| [p(x1)] |
= |
1 · x1
|
| [P(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
r#(p(x1)) |
→ |
p#(p(r(P(x1)))) |
(9) |
|
r#(p(x1)) |
→ |
p#(r(P(x1))) |
(10) |
|
r#(p(x1)) |
→ |
r#(P(x1)) |
(11) |
|
r#(p(x1)) |
→ |
P#(x1) |
(12) |
|
r#(P(P(x1))) |
→ |
P#(P(r(x1))) |
(13) |
|
r#(P(P(x1))) |
→ |
P#(r(x1)) |
(14) |
|
r#(P(P(x1))) |
→ |
r#(x1) |
(15) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.