Certification Problem
Input (TPDB TRS_Standard/SK90/2.52)
The rewrite relation of the following TRS is considered.
|
f(x,0,0) |
→ |
s(x) |
(1) |
|
f(0,y,0) |
→ |
s(y) |
(2) |
|
f(0,0,z) |
→ |
s(z) |
(3) |
|
f(s(0),y,z) |
→ |
f(0,s(y),s(z)) |
(4) |
|
f(s(x),s(y),0) |
→ |
f(x,y,s(0)) |
(5) |
|
f(s(x),0,s(z)) |
→ |
f(x,s(0),z) |
(6) |
|
f(0,s(0),s(0)) |
→ |
s(s(0)) |
(7) |
|
f(s(x),s(y),s(z)) |
→ |
f(x,y,f(s(x),s(y),z)) |
(8) |
|
f(0,s(s(y)),s(0)) |
→ |
f(0,y,s(0)) |
(9) |
|
f(0,s(0),s(s(z))) |
→ |
f(0,s(0),z) |
(10) |
|
f(0,s(s(y)),s(s(z))) |
→ |
f(0,y,f(0,s(s(y)),s(z))) |
(11) |
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) |
= |
0 |
|
stat(0) |
= |
lex
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
| π(f) |
= |
[1,2,3] |
| π(0) |
= |
[] |
| π(s) |
= |
[1] |
all of the following rules can be deleted.
|
f(x,0,0) |
→ |
s(x) |
(1) |
|
f(0,y,0) |
→ |
s(y) |
(2) |
|
f(0,0,z) |
→ |
s(z) |
(3) |
|
f(s(0),y,z) |
→ |
f(0,s(y),s(z)) |
(4) |
|
f(s(x),s(y),0) |
→ |
f(x,y,s(0)) |
(5) |
|
f(s(x),0,s(z)) |
→ |
f(x,s(0),z) |
(6) |
|
f(0,s(0),s(0)) |
→ |
s(s(0)) |
(7) |
|
f(s(x),s(y),s(z)) |
→ |
f(x,y,f(s(x),s(y),z)) |
(8) |
|
f(0,s(s(y)),s(0)) |
→ |
f(0,y,s(0)) |
(9) |
|
f(0,s(0),s(s(z))) |
→ |
f(0,s(0),z) |
(10) |
|
f(0,s(s(y)),s(s(z))) |
→ |
f(0,y,f(0,s(s(y)),s(z))) |
(11) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.