Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z120)
The rewrite relation of the following TRS is considered.
|
c(c(c(a(x1)))) |
→ |
d(d(x1)) |
(1) |
|
d(b(x1)) |
→ |
c(c(x1)) |
(2) |
|
b(c(x1)) |
→ |
b(a(c(x1))) |
(3) |
|
c(x1) |
→ |
a(a(x1)) |
(4) |
|
d(x1) |
→ |
b(c(x1)) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(c(c(c(x1)))) |
→ |
d(d(x1)) |
(6) |
|
b(d(x1)) |
→ |
c(c(x1)) |
(7) |
|
c(b(x1)) |
→ |
c(a(b(x1))) |
(8) |
|
c(x1) |
→ |
a(a(x1)) |
(4) |
|
d(x1) |
→ |
c(b(x1)) |
(9) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
1 · x1 + 2 |
| [d(x1)] |
= |
1 · x1 + 3 |
| [b(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(c(c(c(x1)))) |
→ |
d#(d(x1)) |
(10) |
|
a#(c(c(c(x1)))) |
→ |
d#(x1) |
(11) |
|
b#(d(x1)) |
→ |
c#(c(x1)) |
(12) |
|
b#(d(x1)) |
→ |
c#(x1) |
(13) |
|
c#(b(x1)) |
→ |
c#(a(b(x1))) |
(14) |
|
c#(b(x1)) |
→ |
a#(b(x1)) |
(15) |
|
d#(x1) |
→ |
c#(b(x1)) |
(16) |
|
d#(x1) |
→ |
b#(x1) |
(17) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.