Certification Problem
Input (TPDB TRS_Standard/CiME_04/boolean_rings)
The rewrite relation of the following TRS is considered.
|
xor(x,F) |
→ |
x |
(1) |
|
xor(x,neg(x)) |
→ |
F |
(2) |
|
and(x,T) |
→ |
x |
(3) |
|
and(x,F) |
→ |
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(x,T)) |
(8) |
|
or(x,y) |
→ |
xor(and(x,y),xor(x,y)) |
(9) |
|
equiv(x,y) |
→ |
xor(x,xor(y,T)) |
(10) |
|
neg(x) |
→ |
xor(x,T) |
(11) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(equiv) |
= |
1 |
|
status(equiv) |
= |
[2, 1] |
|
list-extension(equiv) |
= |
Lex |
| prec(or) |
= |
0 |
|
status(or) |
= |
[2, 1] |
|
list-extension(or) |
= |
Lex |
| prec(impl) |
= |
0 |
|
status(impl) |
= |
[2, 1] |
|
list-extension(impl) |
= |
Lex |
| prec(and) |
= |
1 |
|
status(and) |
= |
[1, 2] |
|
list-extension(and) |
= |
Lex |
| prec(T) |
= |
0 |
|
status(T) |
= |
[] |
|
list-extension(T) |
= |
Lex |
| prec(neg) |
= |
0 |
|
status(neg) |
= |
[1] |
|
list-extension(neg) |
= |
Lex |
| prec(xor) |
= |
0 |
|
status(xor) |
= |
[2, 1] |
|
list-extension(xor) |
= |
Lex |
| prec(F) |
= |
0 |
|
status(F) |
= |
[] |
|
list-extension(F) |
= |
Lex |
and the following
Max-polynomial interpretation
| [equiv(x1, x2)] |
=
|
max(4, 1 + 1 · x1, 5 + 1 · x2) |
| [or(x1, x2)] |
=
|
7 + 1 · x1 + 1 · x2
|
| [impl(x1, x2)] |
=
|
7 + 1 · x1 + 1 · x2
|
| [and(x1, x2)] |
=
|
0 + 1 · x1 + 1 · x2
|
| [T] |
=
|
max(1) |
| [neg(x1)] |
=
|
max(0, 2 + 1 · x1) |
| [xor(x1, x2)] |
=
|
max(1, 1 + 1 · x1, 0 + 1 · x2) |
| [F] |
=
|
max(0) |
all of the following rules can be deleted.
|
xor(x,F) |
→ |
x |
(1) |
|
xor(x,neg(x)) |
→ |
F |
(2) |
|
and(x,T) |
→ |
x |
(3) |
|
and(x,F) |
→ |
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(x,T)) |
(8) |
|
or(x,y) |
→ |
xor(and(x,y),xor(x,y)) |
(9) |
|
equiv(x,y) |
→ |
xor(x,xor(y,T)) |
(10) |
|
neg(x) |
→ |
xor(x,T) |
(11) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.