Certification Problem
Input (TPDB TRS_Standard/SK90/2.31)
The rewrite relation of the following TRS is considered.
|
not(true) |
→ |
false |
(1) |
|
not(false) |
→ |
true |
(2) |
|
odd(0) |
→ |
false |
(3) |
|
odd(s(x)) |
→ |
not(odd(x)) |
(4) |
|
+(x,0) |
→ |
x |
(5) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(6) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(true) |
= |
3 |
|
weight(true) |
= |
2 |
|
|
|
| prec(false) |
= |
0 |
|
weight(false) |
= |
2 |
|
|
|
| prec(0) |
= |
1 |
|
weight(0) |
= |
1 |
|
|
|
| prec(not) |
= |
2 |
|
weight(not) |
= |
1 |
|
|
|
| prec(odd) |
= |
4 |
|
weight(odd) |
= |
2 |
|
|
|
| prec(s) |
= |
5 |
|
weight(s) |
= |
1 |
|
|
|
| prec(+) |
= |
6 |
|
weight(+) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
not(true) |
→ |
false |
(1) |
|
not(false) |
→ |
true |
(2) |
|
odd(0) |
→ |
false |
(3) |
|
odd(s(x)) |
→ |
not(odd(x)) |
(4) |
|
+(x,0) |
→ |
x |
(5) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(6) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(7) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.