Certification Problem
Input (TPDB TRS_Standard/SK90/2.58)
The rewrite relation of the following TRS is considered.
|
f(x,y) |
→ |
x |
(1) |
|
g(a) |
→ |
h(a,b,a) |
(2) |
|
i(x) |
→ |
f(x,x) |
(3) |
|
h(x,x,y) |
→ |
g(x) |
(4) |
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
| [h(x1, x2, x3)] |
= |
1 · x1 + 6 · x2 + 2 · x3 + 24 |
| [a] |
= |
15 |
| [b] |
= |
3 |
| [i(x1)] |
= |
16 · x1 + 0 |
| [f(x1, x2)] |
= |
15 · x1 + 1 · x2 + 0 |
| [g(x1)] |
= |
7 · x1 + 24 |
all of the following rules can be deleted.
1.1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(i) |
= |
1 |
|
status(i) |
= |
[1] |
|
list-extension(i) |
= |
Lex |
| prec(h) |
= |
0 |
|
status(h) |
= |
[1, 3, 2] |
|
list-extension(h) |
= |
Lex |
| prec(g) |
= |
0 |
|
status(g) |
= |
[1] |
|
list-extension(g) |
= |
Lex |
| prec(f) |
= |
0 |
|
status(f) |
= |
[1, 2] |
|
list-extension(f) |
= |
Lex |
and the following
Max-polynomial interpretation
| [i(x1)] |
=
|
max(1, 2 + 1 · x1) |
| [h(x1, x2, x3)] |
=
|
max(0, 3 + 1 · x1, 4 + 1 · x2, 1 + 1 · x3) |
| [g(x1)] |
=
|
max(0, 1 + 1 · x1) |
| [f(x1, x2)] |
=
|
max(0, 2 + 1 · x1, 0 + 1 · x2) |
all of the following rules can be deleted.
|
f(x,y) |
→ |
x |
(1) |
|
i(x) |
→ |
f(x,x) |
(3) |
|
h(x,x,y) |
→ |
g(x) |
(4) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.