Certification Problem
Input (TPDB TRS_Standard/SK90/4.32)
The rewrite relation of the following TRS is considered.
|
a(b(x)) |
→ |
b(a(a(x))) |
(1) |
|
b(c(x)) |
→ |
c(b(b(x))) |
(2) |
|
c(a(x)) |
→ |
a(c(c(x))) |
(3) |
|
u(a(x)) |
→ |
x |
(4) |
|
v(b(x)) |
→ |
x |
(5) |
|
w(c(x)) |
→ |
x |
(6) |
|
a(u(x)) |
→ |
x |
(7) |
|
b(v(x)) |
→ |
x |
(8) |
|
c(w(x)) |
→ |
x |
(9) |
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
|
| [b(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
1 · x1
|
| [u(x1)] |
= |
1 · x1 + 1 |
| [v(x1)] |
= |
1 · x1
|
| [w(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
u(a(x)) |
→ |
x |
(4) |
|
a(u(x)) |
→ |
x |
(7) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a(x1)] |
= |
1 · x1
|
| [b(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
1 · x1
|
| [v(x1)] |
= |
1 · x1 + 1 |
| [w(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
v(b(x)) |
→ |
x |
(5) |
|
b(v(x)) |
→ |
x |
(8) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a(x1)] |
= |
1 · x1
|
| [b(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
1 · x1
|
| [w(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
w(c(x)) |
→ |
x |
(6) |
|
c(w(x)) |
→ |
x |
(9) |
1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(b(x)) |
→ |
b#(a(a(x))) |
(10) |
|
a#(b(x)) |
→ |
a#(a(x)) |
(11) |
|
a#(b(x)) |
→ |
a#(x) |
(12) |
|
b#(c(x)) |
→ |
c#(b(b(x))) |
(13) |
|
b#(c(x)) |
→ |
b#(b(x)) |
(14) |
|
b#(c(x)) |
→ |
b#(x) |
(15) |
|
c#(a(x)) |
→ |
a#(c(c(x))) |
(16) |
|
c#(a(x)) |
→ |
c#(c(x)) |
(17) |
|
c#(a(x)) |
→ |
c#(x) |
(18) |
1.1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [a#(x1)] |
= |
0 |
| [b(x1)] |
= |
1 · x1
|
| [b#(x1)] |
= |
1 · x1
|
| [a(x1)] |
= |
0 |
| [c(x1)] |
= |
1 + 1 · x1
|
| [c#(x1)] |
= |
0 |
the
pairs
|
b#(c(x)) |
→ |
c#(b(b(x))) |
(13) |
|
b#(c(x)) |
→ |
b#(b(x)) |
(14) |
|
b#(c(x)) |
→ |
b#(x) |
(15) |
could be deleted.
1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.