Certification Problem
Input (TPDB TRS_Standard/SK90/2.51)
The rewrite relation of the following TRS is considered.
|
ack(0,y) |
→ |
s(y) |
(1) |
|
ack(s(x),0) |
→ |
ack(x,s(0)) |
(2) |
|
ack(s(x),s(y)) |
→ |
ack(x,ack(s(x),y)) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(ack) |
= |
2 |
|
stat(ack) |
= |
lex
|
| prec(0) |
= |
0 |
|
stat(0) |
= |
lex
|
| prec(s) |
= |
1 |
|
stat(s) |
= |
lex
|
| π(ack) |
= |
[1,2] |
| π(0) |
= |
[] |
| π(s) |
= |
[1] |
all of the following rules can be deleted.
|
ack(0,y) |
→ |
s(y) |
(1) |
|
ack(s(x),0) |
→ |
ack(x,s(0)) |
(2) |
|
ack(s(x),s(y)) |
→ |
ack(x,ack(s(x),y)) |
(3) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.