Certification Problem
Input (TPDB TRS_Standard/CiME_04/ack_prolog)
The rewrite relation of the following TRS is considered.
|
ack_in(0,n) |
→ |
ack_out(s(n)) |
(1) |
|
ack_in(s(m),0) |
→ |
u11(ack_in(m,s(0))) |
(2) |
|
u11(ack_out(n)) |
→ |
ack_out(n) |
(3) |
|
ack_in(s(m),s(n)) |
→ |
u21(ack_in(s(m),n),m) |
(4) |
|
u21(ack_out(n),m) |
→ |
u22(ack_in(m,n)) |
(5) |
|
u22(ack_out(n)) |
→ |
ack_out(n) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(ack_in) |
= |
2 |
|
stat(ack_in) |
= |
lex
|
| prec(0) |
= |
1 |
|
stat(0) |
= |
lex
|
| prec(ack_out) |
= |
1 |
|
stat(ack_out) |
= |
lex
|
| prec(s) |
= |
1 |
|
stat(s) |
= |
lex
|
| prec(u11) |
= |
0 |
|
stat(u11) |
= |
lex
|
| prec(u21) |
= |
2 |
|
stat(u21) |
= |
lex
|
| π(ack_in) |
= |
[1,2] |
| π(0) |
= |
[] |
| π(ack_out) |
= |
[1] |
| π(s) |
= |
[1] |
| π(u11) |
= |
[1] |
| π(u21) |
= |
[2,1] |
| π(u22) |
= |
1 |
all of the following rules can be deleted.
|
ack_in(0,n) |
→ |
ack_out(s(n)) |
(1) |
|
ack_in(s(m),0) |
→ |
u11(ack_in(m,s(0))) |
(2) |
|
u11(ack_out(n)) |
→ |
ack_out(n) |
(3) |
|
ack_in(s(m),s(n)) |
→ |
u21(ack_in(s(m),n),m) |
(4) |
|
u21(ack_out(n),m) |
→ |
u22(ack_in(m,n)) |
(5) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(u22) |
= |
1 |
|
weight(u22) |
= |
0 |
|
|
|
| prec(ack_out) |
= |
0 |
|
weight(ack_out) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
u22(ack_out(n)) |
→ |
ack_out(n) |
(6) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.