Certification Problem
Input (TPDB TRS_Standard/SK90/2.16)
The rewrite relation of the following TRS is considered.
|
f(0) |
→ |
1 |
(1) |
|
f(s(x)) |
→ |
g(x,s(x)) |
(2) |
|
g(0,y) |
→ |
y |
(3) |
|
g(s(x),y) |
→ |
g(x,+(y,s(x))) |
(4) |
|
+(x,0) |
→ |
x |
(5) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(6) |
|
g(s(x),y) |
→ |
g(x,s(+(y,x))) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(f) |
= |
2 |
|
stat(f) |
= |
lex
|
| prec(0) |
= |
3 |
|
stat(0) |
= |
lex
|
| prec(1) |
= |
3 |
|
stat(1) |
= |
lex
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
| prec(g) |
= |
2 |
|
stat(g) |
= |
lex
|
| prec(+) |
= |
1 |
|
stat(+) |
= |
lex
|
| π(f) |
= |
[1] |
| π(0) |
= |
[] |
| π(1) |
= |
[] |
| π(s) |
= |
[1] |
| π(g) |
= |
[1,2] |
| π(+) |
= |
[2,1] |
all of the following rules can be deleted.
|
f(0) |
→ |
1 |
(1) |
|
f(s(x)) |
→ |
g(x,s(x)) |
(2) |
|
g(0,y) |
→ |
y |
(3) |
|
g(s(x),y) |
→ |
g(x,+(y,s(x))) |
(4) |
|
+(x,0) |
→ |
x |
(5) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(6) |
|
g(s(x),y) |
→ |
g(x,s(+(y,x))) |
(7) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.