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 AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [and(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [false] |
= |
0 |
| [not(x1)] |
= |
2 · x1
|
| [implies(x1, x2)] |
= |
1 + 2 · x1 + 2 · x2
|
all of the following rules can be deleted.
|
implies(false,y) |
→ |
not(false) |
(4) |
|
implies(x,false) |
→ |
not(x) |
(5) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [and(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [false] |
= |
0 |
| [not(x1)] |
= |
1 + 2 · x1
|
| [implies(x1, x2)] |
= |
1 · x1 + 2 · x2
|
all of the following rules can be deleted.
|
and(x,not(false)) |
→ |
x |
(2) |
|
not(not(x)) |
→ |
x |
(3) |
|
implies(not(x),not(y)) |
→ |
implies(y,and(x,y)) |
(6) |
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(false) |
= |
1 |
|
weight(false) |
= |
1 |
|
|
|
| prec(and) |
= |
0 |
|
weight(and) |
= |
0 |
|
|
|
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.