Certification Problem
Input (TPDB TRS_Standard/SK90/4.53)
The rewrite relation of the following TRS is considered.
|
f(a) |
→ |
b |
(1) |
|
f(c) |
→ |
d |
(2) |
|
f(g(x,y)) |
→ |
g(f(x),f(y)) |
(3) |
|
f(h(x,y)) |
→ |
g(h(y,f(x)),h(x,f(y))) |
(4) |
|
g(x,x) |
→ |
h(e,x) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(f) |
= |
4 |
|
stat(f) |
= |
mul
|
| prec(a) |
= |
5 |
|
stat(a) |
= |
mul
|
| prec(b) |
= |
1 |
|
stat(b) |
= |
mul
|
| prec(c) |
= |
7 |
|
stat(c) |
= |
mul
|
| prec(d) |
= |
6 |
|
stat(d) |
= |
mul
|
| prec(g) |
= |
3 |
|
stat(g) |
= |
mul
|
| prec(h) |
= |
0 |
|
stat(h) |
= |
mul
|
| prec(e) |
= |
2 |
|
stat(e) |
= |
mul
|
| π(f) |
= |
[1] |
| π(a) |
= |
[] |
| π(b) |
= |
[] |
| π(c) |
= |
[] |
| π(d) |
= |
[] |
| π(g) |
= |
[1,2] |
| π(h) |
= |
[1,2] |
| π(e) |
= |
[] |
all of the following rules can be deleted.
|
f(a) |
→ |
b |
(1) |
|
f(c) |
→ |
d |
(2) |
|
f(g(x,y)) |
→ |
g(f(x),f(y)) |
(3) |
|
f(h(x,y)) |
→ |
g(h(y,f(x)),h(x,f(y))) |
(4) |
|
g(x,x) |
→ |
h(e,x) |
(5) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.