Certification Problem
Input (TPDB TRS_Standard/SK90/2.15)
The rewrite relation of the following TRS is considered.
|
f(0) |
→ |
1 |
(1) |
|
f(s(x)) |
→ |
g(f(x)) |
(2) |
|
g(x) |
→ |
+(x,s(x)) |
(3) |
|
f(s(x)) |
→ |
+(f(x),s(f(x))) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(f) |
= |
4 |
|
stat(f) |
= |
lex
|
| prec(0) |
= |
5 |
|
stat(0) |
= |
lex
|
| prec(1) |
= |
1 |
|
stat(1) |
= |
lex
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
| prec(g) |
= |
3 |
|
stat(g) |
= |
lex
|
| prec(+) |
= |
2 |
|
stat(+) |
= |
lex
|
| π(f) |
= |
[1] |
| π(0) |
= |
[] |
| π(1) |
= |
[] |
| π(s) |
= |
[1] |
| π(g) |
= |
[1] |
| π(+) |
= |
[1,2] |
all of the following rules can be deleted.
|
f(0) |
→ |
1 |
(1) |
|
f(s(x)) |
→ |
g(f(x)) |
(2) |
|
g(x) |
→ |
+(x,s(x)) |
(3) |
|
f(s(x)) |
→ |
+(f(x),s(f(x))) |
(4) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.