Certification Problem
Input (TPDB TRS_Standard/SK90/2.35)
The rewrite relation of the following TRS is considered.
|
and(x,false) |
→ |
false |
(1) |
|
and(x,not(false)) |
→ |
x |
(2) |
|
not(not(x)) |
→ |
x |
(3) |
|
implies(false,y) |
→ |
not(false) |
(4) |
|
implies(x,false) |
→ |
not(x) |
(5) |
|
implies(not(x),not(y)) |
→ |
implies(y,and(x,y)) |
(6) |
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
| [and(x1, x2)] |
= |
· x1 + · x2 +
|
| [implies(x1, x2)] |
= |
· x1 + · x2 +
|
| [false] |
= |
|
| [not(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
and(x,false) |
→ |
false |
(1) |
|
and(x,not(false)) |
→ |
x |
(2) |
|
implies(x,false) |
→ |
not(x) |
(5) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [and(x1, x2)] |
= |
· x1 + · x2 +
|
| [implies(x1, x2)] |
= |
· x1 + · x2 +
|
| [false] |
= |
|
| [not(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
not(not(x)) |
→ |
x |
(3) |
|
implies(not(x),not(y)) |
→ |
implies(y,and(x,y)) |
(6) |
1.1.1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(implies) |
= |
1 |
|
status(implies) |
= |
[1, 2] |
|
list-extension(implies) |
= |
Lex |
| prec(not) |
= |
0 |
|
status(not) |
= |
[1] |
|
list-extension(not) |
= |
Lex |
| prec(false) |
= |
0 |
|
status(false) |
= |
[] |
|
list-extension(false) |
= |
Lex |
and the following
Max-polynomial interpretation
| [implies(x1, x2)] |
=
|
max(4, 0 + 1 · x1, 0 + 1 · x2) |
| [not(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [false] |
=
|
max(4) |
all of the following rules can be deleted.
|
implies(false,y) |
→ |
not(false) |
(4) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.