Certification Problem
Input (TPDB TRS_Standard/SK90/4.16)
The rewrite relation of the following TRS is considered.
|
f(0) |
→ |
s(0) |
(1) |
|
f(s(0)) |
→ |
s(s(0)) |
(2) |
|
f(s(0)) |
→ |
*(s(s(0)),f(0)) |
(3) |
|
f(+(x,s(0))) |
→ |
+(s(s(0)),f(x)) |
(4) |
|
f(+(x,y)) |
→ |
*(f(x),f(y)) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [+(x1, x2)] |
= |
8 · x1 + 8 · x2 + 1 |
| [f(x1)] |
= |
8 · x1 + 0 |
| [*(x1, x2)] |
= |
8 · x1 + 8 · x2 + 0 |
| [0] |
= |
0 |
| [s(x1)] |
= |
4 · x1 + 0 |
all of the following rules can be deleted.
|
f(+(x,s(0))) |
→ |
+(s(s(0)),f(x)) |
(4) |
|
f(+(x,y)) |
→ |
*(f(x),f(y)) |
(5) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1
over the naturals
| [f(x1)] |
= |
|
| 1 |
1 |
0 |
0 |
| 1 |
1 |
1 |
0 |
| 0 |
1 |
0 |
0 |
| 1 |
1 |
1 |
0 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
|
|
|
| [*(x1, x2)] |
= |
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
1 |
0 |
| 0 |
0 |
0 |
1 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
| 0 |
1 |
1 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
|
|
· x2 +
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [0] |
= |
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
|
|
|
| [s(x1)] |
= |
|
| 1 |
0 |
0 |
1 |
| 0 |
1 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 1 |
1 |
0 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1
over the naturals
| [f(x1)] |
= |
|
| 1 |
1 |
0 |
1 |
| 1 |
0 |
1 |
1 |
| 0 |
1 |
0 |
1 |
| 0 |
1 |
0 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [*(x1, x2)] |
= |
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
| 0 |
1 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
· x2 +
|
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [0] |
= |
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
|
|
|
| [s(x1)] |
= |
|
| 1 |
0 |
0 |
1 |
| 1 |
0 |
1 |
1 |
| 0 |
0 |
0 |
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.
|
f(s(0)) |
→ |
*(s(s(0)),f(0)) |
(3) |
1.1.1.1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(s) |
= |
0 |
|
status(s) |
= |
[1] |
|
list-extension(s) |
= |
Lex |
| prec(f) |
= |
0 |
|
status(f) |
= |
[1] |
|
list-extension(f) |
= |
Lex |
| prec(0) |
= |
0 |
|
status(0) |
= |
[] |
|
list-extension(0) |
= |
Lex |
and the following
Max-polynomial interpretation
| [s(x1)] |
=
|
max(0, 1 + 1 · x1) |
| [f(x1)] |
=
|
max(1, 2 + 1 · x1) |
| [0] |
=
|
max(1) |
all of the following rules can be deleted.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.