Certification Problem
Input (TPDB TRS_Standard/SK90/4.20)
The rewrite relation of the following TRS is considered.
|
not(x) |
→ |
xor(x,true) |
(1) |
|
or(x,y) |
→ |
xor(and(x,y),xor(x,y)) |
(2) |
|
implies(x,y) |
→ |
xor(and(x,y),xor(x,true)) |
(3) |
|
and(x,true) |
→ |
x |
(4) |
|
and(x,false) |
→ |
false |
(5) |
|
and(x,x) |
→ |
x |
(6) |
|
xor(x,false) |
→ |
x |
(7) |
|
xor(x,x) |
→ |
false |
(8) |
|
and(xor(x,y),z) |
→ |
xor(and(x,z),and(y,z)) |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(not) |
= |
1 |
|
stat(not) |
= |
mul
|
| prec(xor) |
= |
0 |
|
stat(xor) |
= |
mul
|
| prec(true) |
= |
1 |
|
stat(true) |
= |
mul
|
| prec(or) |
= |
3 |
|
stat(or) |
= |
mul
|
| prec(and) |
= |
2 |
|
stat(and) |
= |
mul
|
| prec(implies) |
= |
4 |
|
stat(implies) |
= |
mul
|
| prec(false) |
= |
0 |
|
stat(false) |
= |
mul
|
| π(not) |
= |
[1] |
| π(xor) |
= |
[1,2] |
| π(true) |
= |
[] |
| π(or) |
= |
[1,2] |
| π(and) |
= |
[1,2] |
| π(implies) |
= |
[1,2] |
| π(false) |
= |
[] |
all of the following rules can be deleted.
|
not(x) |
→ |
xor(x,true) |
(1) |
|
or(x,y) |
→ |
xor(and(x,y),xor(x,y)) |
(2) |
|
implies(x,y) |
→ |
xor(and(x,y),xor(x,true)) |
(3) |
|
and(x,true) |
→ |
x |
(4) |
|
and(x,false) |
→ |
false |
(5) |
|
and(x,x) |
→ |
x |
(6) |
|
xor(x,false) |
→ |
x |
(7) |
|
xor(x,x) |
→ |
false |
(8) |
|
and(xor(x,y),z) |
→ |
xor(and(x,z),and(y,z)) |
(9) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.