Certification Problem
Input (TPDB TRS_Standard/SK90/2.07)
The rewrite relation of the following TRS is considered.
|
f(0,y) |
→ |
y |
(1) |
|
f(x,0) |
→ |
x |
(2) |
|
f(i(x),y) |
→ |
i(x) |
(3) |
|
f(f(x,y),z) |
→ |
f(x,f(y,z)) |
(4) |
|
f(g(x,y),z) |
→ |
g(f(x,z),f(y,z)) |
(5) |
|
f(1,g(x,y)) |
→ |
x |
(6) |
|
f(2,g(x,y)) |
→ |
y |
(7) |
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) |
= |
2 |
|
stat(0) |
= |
lex
|
| prec(g) |
= |
0 |
|
stat(g) |
= |
lex
|
| prec(1) |
= |
3 |
|
stat(1) |
= |
lex
|
| prec(2) |
= |
4 |
|
stat(2) |
= |
lex
|
| π(f) |
= |
[1,2] |
| π(0) |
= |
[] |
| π(i) |
= |
1 |
| π(g) |
= |
[1,2] |
| π(1) |
= |
[] |
| π(2) |
= |
[] |
all of the following rules can be deleted.
|
f(0,y) |
→ |
y |
(1) |
|
f(x,0) |
→ |
x |
(2) |
|
f(i(x),y) |
→ |
i(x) |
(3) |
|
f(f(x,y),z) |
→ |
f(x,f(y,z)) |
(4) |
|
f(g(x,y),z) |
→ |
g(f(x,z),f(y,z)) |
(5) |
|
f(1,g(x,y)) |
→ |
x |
(6) |
|
f(2,g(x,y)) |
→ |
y |
(7) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.