Certification Problem
Input (TPDB TRS_Relative/INVY_15/invNSS03)
The relative rewrite relation R/S is considered where R is the following TRS
|
f(x,0) |
→ |
s(x) |
(1) |
|
g(x) |
→ |
h(x,gen) |
(2) |
|
h(0,x) |
→ |
f(x,x) |
(3) |
| a |
→ |
b |
(4) |
and S is the following TRS.
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(f) |
= |
1 |
|
stat(f) |
= |
lex
|
| prec(0) |
= |
2 |
|
stat(0) |
= |
lex
|
| prec(g) |
= |
4 |
|
stat(g) |
= |
lex
|
| prec(h) |
= |
3 |
|
stat(h) |
= |
lex
|
| prec(gen) |
= |
0 |
|
stat(gen) |
= |
lex
|
| prec(a) |
= |
6 |
|
stat(a) |
= |
lex
|
| prec(b) |
= |
5 |
|
stat(b) |
= |
lex
|
| π(f) |
= |
[2,1] |
| π(0) |
= |
[] |
| π(s) |
= |
1 |
| π(g) |
= |
[1] |
| π(h) |
= |
[2,1] |
| π(gen) |
= |
[] |
| π(a) |
= |
[] |
| π(b) |
= |
[] |
all of the following rules can be deleted.
|
f(x,0) |
→ |
s(x) |
(1) |
|
g(x) |
→ |
h(x,gen) |
(2) |
|
h(0,x) |
→ |
f(x,x) |
(3) |
| a |
→ |
b |
(4) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.