Certification Problem
Input (TPDB SRS_Standard/Zantema_06/beans5)
The rewrite relation of the following TRS is considered.
|
b(a(a(x1))) |
→ |
a(b(c(x1))) |
(1) |
|
c(a(x1)) |
→ |
a(c(x1)) |
(2) |
|
c(b(x1)) |
→ |
b(a(x1)) |
(3) |
|
a(a(x1)) |
→ |
a(b(a(x1))) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(a(a(x1))) |
→ |
a#(b(c(x1))) |
(5) |
|
b#(a(a(x1))) |
→ |
b#(c(x1)) |
(6) |
|
b#(a(a(x1))) |
→ |
c#(x1) |
(7) |
|
c#(a(x1)) |
→ |
a#(c(x1)) |
(8) |
|
c#(a(x1)) |
→ |
c#(x1) |
(9) |
|
c#(b(x1)) |
→ |
b#(a(x1)) |
(10) |
|
c#(b(x1)) |
→ |
a#(x1) |
(11) |
|
a#(a(x1)) |
→ |
a#(b(a(x1))) |
(12) |
|
a#(a(x1)) |
→ |
b#(a(x1)) |
(13) |
1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [b#(x1)] |
= |
1 · x1
|
| [a(x1)] |
= |
1 + 1 · x1
|
| [a#(x1)] |
= |
1 · x1
|
| [b(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
1 + 1 · x1
|
| [c#(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
b#(a(a(x1))) |
→ |
a#(b(c(x1))) |
(5) |
|
b#(a(a(x1))) |
→ |
b#(c(x1)) |
(6) |
|
b#(a(a(x1))) |
→ |
c#(x1) |
(7) |
|
c#(a(x1)) |
→ |
a#(c(x1)) |
(8) |
|
c#(a(x1)) |
→ |
c#(x1) |
(9) |
|
c#(b(x1)) |
→ |
a#(x1) |
(11) |
could be deleted.
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.