Certification Problem
Input (TPDB TRS_Standard/SK90/2.30)
The rewrite relation of the following TRS is considered.
|
not(x) |
→ |
xor(x,true) |
(1) |
|
implies(x,y) |
→ |
xor(and(x,y),xor(x,true)) |
(2) |
|
or(x,y) |
→ |
xor(and(x,y),xor(x,y)) |
(3) |
|
=(x,y) |
→ |
xor(x,xor(y,true)) |
(4) |
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
| [not(x1)] |
= |
1 + 1 · x1
|
| [xor(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [true] |
= |
1 |
| [implies(x1, x2)] |
= |
2 + 2 · x1 + 1 · x2
|
| [and(x1, x2)] |
= |
1 + 1 · x1 + 1 · x2
|
| [or(x1, x2)] |
= |
1 + 2 · x1 + 2 · x2
|
| [=(x1, x2)] |
= |
2 + 2 · x1 + 2 · x2
|
all of the following rules can be deleted.
|
=(x,y) |
→ |
xor(x,xor(y,true)) |
(4) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [not(x1)] |
= |
1 + 2 · x1
|
| [xor(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [true] |
= |
0 |
| [implies(x1, x2)] |
= |
2 + 2 · x1 + 1 · x2
|
| [and(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [or(x1, x2)] |
= |
2 · x1 + 2 · x2
|
all of the following rules can be deleted.
|
not(x) |
→ |
xor(x,true) |
(1) |
|
implies(x,y) |
→ |
xor(and(x,y),xor(x,true)) |
(2) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [or(x1, x2)] |
= |
2 + 2 · x1 + 2 · x2
|
| [xor(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [and(x1, x2)] |
= |
1 + 1 · x1 + 1 · x2
|
all of the following rules can be deleted.
|
or(x,y) |
→ |
xor(and(x,y),xor(x,y)) |
(3) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.