Certification Problem
Input (TPDB TRS_Standard/Secret_06_TRS/2)
The rewrite relation of the following TRS is considered.
|
c(c(c(b(x)))) |
→ |
a(1,b(c(x))) |
(1) |
|
b(c(b(c(x)))) |
→ |
a(0,a(1,x)) |
(2) |
|
a(0,x) |
→ |
c(c(x)) |
(3) |
|
a(1,x) |
→ |
c(b(x)) |
(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.
|
c#(c(c(b(x)))) |
→ |
a#(1,b(c(x))) |
(5) |
|
c#(c(c(b(x)))) |
→ |
b#(c(x)) |
(6) |
|
c#(c(c(b(x)))) |
→ |
c#(x) |
(7) |
|
b#(c(b(c(x)))) |
→ |
a#(0,a(1,x)) |
(8) |
|
b#(c(b(c(x)))) |
→ |
a#(1,x) |
(9) |
|
a#(0,x) |
→ |
c#(c(x)) |
(10) |
|
a#(0,x) |
→ |
c#(x) |
(11) |
|
a#(1,x) |
→ |
c#(b(x)) |
(12) |
|
a#(1,x) |
→ |
b#(x) |
(13) |
1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [c(x1)] |
= |
1 + 1 · x1
|
| [b(x1)] |
= |
1 + 1 · x1
|
| [a(x1, x2)] |
= |
2 + 2 · x1 + 1 · x2
|
| [1] |
= |
0 |
| [0] |
= |
0 |
| [c#(x1)] |
= |
1 · x1
|
| [a#(x1, x2)] |
= |
1 + 1 · x1 + 1 · x2
|
| [b#(x1)] |
= |
1 · x1
|
the
pairs
|
c#(c(c(b(x)))) |
→ |
b#(c(x)) |
(6) |
|
c#(c(c(b(x)))) |
→ |
c#(x) |
(7) |
|
b#(c(b(c(x)))) |
→ |
a#(1,x) |
(9) |
|
a#(0,x) |
→ |
c#(x) |
(11) |
|
a#(1,x) |
→ |
b#(x) |
(13) |
and
no rules
could be deleted.
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.