Certification Problem
Input (TPDB TRS_Standard/SK90/4.21)
The rewrite relation of the following TRS is considered.
|
and(x,or(y,z)) |
→ |
or(and(x,y),and(x,z)) |
(1) |
|
and(x,and(y,y)) |
→ |
and(x,y) |
(2) |
|
or(or(x,y),and(y,z)) |
→ |
or(x,y) |
(3) |
|
or(x,and(x,y)) |
→ |
x |
(4) |
|
or(true,y) |
→ |
true |
(5) |
|
or(x,false) |
→ |
x |
(6) |
|
or(x,x) |
→ |
x |
(7) |
|
or(x,or(y,y)) |
→ |
or(x,y) |
(8) |
|
and(x,true) |
→ |
x |
(9) |
|
and(false,y) |
→ |
false |
(10) |
|
and(x,x) |
→ |
x |
(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(false) |
= |
0 |
|
status(false) |
= |
[] |
|
list-extension(false) |
= |
Lex |
| prec(true) |
= |
0 |
|
status(true) |
= |
[] |
|
list-extension(true) |
= |
Lex |
| prec(and) |
= |
1 |
|
status(and) |
= |
[1, 2] |
|
list-extension(and) |
= |
Lex |
| prec(or) |
= |
0 |
|
status(or) |
= |
[1, 2] |
|
list-extension(or) |
= |
Lex |
and the following
Max-polynomial interpretation
| [false] |
=
|
max(0) |
| [true] |
=
|
max(0) |
| [and(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 7 + 1 · x2) |
| [or(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
all of the following rules can be deleted.
|
and(x,or(y,z)) |
→ |
or(and(x,y),and(x,z)) |
(1) |
|
and(x,and(y,y)) |
→ |
and(x,y) |
(2) |
|
or(or(x,y),and(y,z)) |
→ |
or(x,y) |
(3) |
|
or(x,and(x,y)) |
→ |
x |
(4) |
|
or(true,y) |
→ |
true |
(5) |
|
or(x,false) |
→ |
x |
(6) |
|
or(x,x) |
→ |
x |
(7) |
|
or(x,or(y,y)) |
→ |
or(x,y) |
(8) |
|
and(x,true) |
→ |
x |
(9) |
|
and(false,y) |
→ |
false |
(10) |
|
and(x,x) |
→ |
x |
(11) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.