Certification Problem
Input (TPDB SRS_Relative/Mixed_relative_SRS/zr02)
The relative rewrite relation R/S is considered where R is the following TRS
|
n(s(x1)) |
→ |
s(x1) |
(1) |
|
o(s(x1)) |
→ |
s(x1) |
(2) |
|
n(o(p(x1))) |
→ |
o(n(x1)) |
(3) |
and S is the following TRS.
|
t(x1) |
→ |
t(c(n(x1))) |
(4) |
|
n(p(x1)) |
→ |
p(n(x1)) |
(5) |
|
p(m(x1)) |
→ |
m(p(x1)) |
(6) |
|
p(s(x1)) |
→ |
s(x1) |
(7) |
|
c(m(x1)) |
→ |
m(c(x1)) |
(8) |
|
o(n(x1)) |
→ |
n(o(x1)) |
(9) |
|
p(n(x1)) |
→ |
m(p(x1)) |
(10) |
|
o(m(x1)) |
→ |
n(o(x1)) |
(11) |
|
c(n(x1)) |
→ |
n(c(x1)) |
(12) |
|
c(p(x1)) |
→ |
p(c(x1)) |
(13) |
|
c(o(x1)) |
→ |
o(c(x1)) |
(14) |
|
c(o(x1)) |
→ |
o(x1) |
(15) |
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
| [n(x1)] |
= |
1 · x1
|
| [s(x1)] |
= |
1 · x1
|
| [o(x1)] |
= |
1 + 1 · x1
|
| [p(x1)] |
= |
1 · x1
|
| [t(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
1 · x1
|
| [m(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [n(x1)] |
= |
1 · x1
|
| [s(x1)] |
= |
1 · x1
|
| [o(x1)] |
= |
1 · x1
|
| [p(x1)] |
= |
1 + 1 · x1
|
| [t(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
1 · x1
|
| [m(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
n(o(p(x1))) |
→ |
o(n(x1)) |
(3) |
|
p(s(x1)) |
→ |
s(x1) |
(7) |
1.1.1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
| [n(x1)] |
= |
+ · x1
|
| [s(x1)] |
= |
+ · x1
|
| [t(x1)] |
= |
+ · x1
|
| [c(x1)] |
= |
+ · x1
|
| [p(x1)] |
= |
+ · x1
|
| [m(x1)] |
= |
+ · x1
|
| [o(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.