Certification Problem
Input (TPDB TRS_Relative/INVY_15/#3.15_rand)
The relative rewrite relation R/S is considered where R is the following TRS
average(s(x),y) |
→ |
average(x,s(y)) |
(1) |
average(x,s(s(s(y)))) |
→ |
s(average(s(x),y)) |
(2) |
average(0,0) |
→ |
0 |
(3) |
average(0,s(0)) |
→ |
0 |
(4) |
average(0,s(s(0))) |
→ |
s(0) |
(5) |
and S is the following TRS.
rand(x) |
→ |
x |
(6) |
rand(x) |
→ |
rand(s(x)) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
[average(x1, x2)] |
= |
· x1 + · x2 +
|
[rand(x1)] |
= |
· x1 +
|
[s(x1)] |
= |
· x1 +
|
[0] |
= |
|
all of the following rules can be deleted.
average(0,0) |
→ |
0 |
(3) |
average(0,s(0)) |
→ |
0 |
(4) |
average(0,s(s(0))) |
→ |
s(0) |
(5) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
[average(x1, x2)] |
= |
· x1 + · x2 +
|
[rand(x1)] |
= |
· x1 +
|
[s(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
[average(x1, x2)] |
= |
· x1 + · x2 +
|
[rand(x1)] |
= |
· x1 +
|
[s(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
average(x,s(s(s(y)))) |
→ |
s(average(s(x),y)) |
(2) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1
over the naturals
[average(x1, x2)] |
= |
|
1 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
|
|
· x1 +
|
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
· x2 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
[rand(x1)] |
= |
|
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
[s(x1)] |
= |
|
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
average(s(x),y) |
→ |
average(x,s(y)) |
(1) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.