Certification Problem
Input (TPDB TRS_Standard/SK90/2.40)
The rewrite relation of the following TRS is considered.
|
or(true,y) |
→ |
true |
(1) |
|
or(x,true) |
→ |
true |
(2) |
|
or(false,false) |
→ |
false |
(3) |
|
mem(x,nil) |
→ |
false |
(4) |
|
mem(x,set(y)) |
→ |
=(x,y) |
(5) |
|
mem(x,union(y,z)) |
→ |
or(mem(x,y),mem(x,z)) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(or) |
= |
1 |
|
stat(or) |
= |
lex
|
| prec(true) |
= |
0 |
|
stat(true) |
= |
lex
|
| prec(false) |
= |
0 |
|
stat(false) |
= |
lex
|
| prec(mem) |
= |
2 |
|
stat(mem) |
= |
lex
|
| prec(nil) |
= |
0 |
|
stat(nil) |
= |
lex
|
| prec(=) |
= |
2 |
|
stat(=) |
= |
lex
|
| prec(union) |
= |
3 |
|
stat(union) |
= |
lex
|
| π(or) |
= |
[2,1] |
| π(true) |
= |
[] |
| π(false) |
= |
[] |
| π(mem) |
= |
[1,2] |
| π(nil) |
= |
[] |
| π(set) |
= |
1 |
| π(=) |
= |
[1,2] |
| π(union) |
= |
[2,1] |
all of the following rules can be deleted.
|
or(true,y) |
→ |
true |
(1) |
|
or(x,true) |
→ |
true |
(2) |
|
or(false,false) |
→ |
false |
(3) |
|
mem(x,nil) |
→ |
false |
(4) |
|
mem(x,union(y,z)) |
→ |
or(mem(x,y),mem(x,z)) |
(6) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(set) |
= |
2 |
|
weight(set) |
= |
1 |
|
|
|
| prec(mem) |
= |
1 |
|
weight(mem) |
= |
0 |
|
|
|
| prec(=) |
= |
0 |
|
weight(=) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
mem(x,set(y)) |
→ |
=(x,y) |
(5) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.