Certification Problem
Input (TPDB TRS_Standard/TCT_12/polycounter-5)
The rewrite relation of the following TRS is considered.
|
f(s(x1),x2,x3,x4,x5) |
→ |
f(x1,x2,x3,x4,x5) |
(1) |
|
f(0,s(x2),x3,x4,x5) |
→ |
f(x2,x2,x3,x4,x5) |
(2) |
|
f(0,0,s(x3),x4,x5) |
→ |
f(x3,x3,x3,x4,x5) |
(3) |
|
f(0,0,0,s(x4),x5) |
→ |
f(x4,x4,x4,x4,x5) |
(4) |
|
f(0,0,0,0,s(x5)) |
→ |
f(x5,x5,x5,x5,x5) |
(5) |
|
f(0,0,0,0,0) |
→ |
0 |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(f) |
= |
0 |
|
stat(f) |
= |
lex
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
| prec(0) |
= |
0 |
|
stat(0) |
= |
lex
|
| π(f) |
= |
[5,4,3,2,1] |
| π(s) |
= |
[1] |
| π(0) |
= |
[] |
all of the following rules can be deleted.
|
f(s(x1),x2,x3,x4,x5) |
→ |
f(x1,x2,x3,x4,x5) |
(1) |
|
f(0,s(x2),x3,x4,x5) |
→ |
f(x2,x2,x3,x4,x5) |
(2) |
|
f(0,0,s(x3),x4,x5) |
→ |
f(x3,x3,x3,x4,x5) |
(3) |
|
f(0,0,0,s(x4),x5) |
→ |
f(x4,x4,x4,x4,x5) |
(4) |
|
f(0,0,0,0,s(x5)) |
→ |
f(x5,x5,x5,x5,x5) |
(5) |
|
f(0,0,0,0,0) |
→ |
0 |
(6) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.