Certification Problem
Input (TPDB TRS_Equational/Mixed_AC/boolean_rings)
The rewrite relation of the following equational TRS is considered.
|
xor(F,x) |
→ |
x |
(1) |
|
xor(neg(x),x) |
→ |
F |
(2) |
|
and(T,x) |
→ |
x |
(3) |
|
and(F,x) |
→ |
F |
(4) |
|
and(x,x) |
→ |
x |
(5) |
|
and(xor(x,y),z) |
→ |
xor(and(x,z),and(y,z)) |
(6) |
|
xor(x,x) |
→ |
F |
(7) |
|
impl(x,y) |
→ |
xor(and(x,y),xor(T,x)) |
(8) |
|
or(x,y) |
→ |
xor(and(x,y),xor(x,y)) |
(9) |
|
equiv(x,y) |
→ |
xor(xor(T,y),x) |
(10) |
|
neg(x) |
→ |
xor(T,x) |
(11) |
Associative symbols: and, or, xor
Commutative symbols: and, or, xor
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [and(x1, x2)] |
= |
1 · x2 + 1 · x1 + 2 · x1 · x2
|
| [or(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [xor(x1, x2)] |
= |
1 + 1 · x2 + 1 · x1
|
| [F] |
= |
1 |
| [neg(x1)] |
= |
3 + 3 · x1 + 3 · x1 · x1
|
| [T] |
= |
0 |
| [impl(x1, x2)] |
= |
3 + 1 · x2 + 3 · x1 + 2 · x1 · x2
|
| [equiv(x1, x2)] |
= |
3 + 1 · x2 + 1 · x1
|
the
rules
|
xor(F,x) |
→ |
x |
(1) |
|
xor(neg(x),x) |
→ |
F |
(2) |
|
impl(x,y) |
→ |
xor(and(x,y),xor(T,x)) |
(8) |
|
or(x,y) |
→ |
xor(and(x,y),xor(x,y)) |
(9) |
|
equiv(x,y) |
→ |
xor(xor(T,y),x) |
(10) |
|
neg(x) |
→ |
xor(T,x) |
(11) |
can be deleted.
1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [and(x1, x2)] |
= |
1 · x2 + 1 · x1 + 1 · x1 · x2
|
| [or(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [xor(x1, x2)] |
= |
1 + 1 · x2 + 1 · x1
|
| [T] |
= |
2 |
| [F] |
= |
0 |
the
rules
|
and(T,x) |
→ |
x |
(3) |
|
xor(x,x) |
→ |
F |
(7) |
can be deleted.
1.1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [and(x1, x2)] |
= |
2 + 2 · x2 + 2 · x1 + 1 · x1 · x2
|
| [or(x1, x2)] |
= |
1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
|
| [xor(x1, x2)] |
= |
2 + 1 · x2 + 1 · x1
|
| [F] |
= |
3 |
the
rules
|
and(F,x) |
→ |
F |
(4) |
|
and(x,x) |
→ |
x |
(5) |
can be deleted.
1.1.1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [and(x1, x2)] |
= |
2 + 3 · x2 + 3 · x1 + 3 · x1 · x2
|
| [or(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [xor(x1, x2)] |
= |
2 + 1 · x2 + 1 · x1
|
the
rule
|
and(xor(x,y),z) |
→ |
xor(and(x,z),and(y,z)) |
(6) |
can be deleted.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is AC-terminating.