Certification Problem
Input (TPDB TRS_Standard/SK90/4.56)
The rewrite relation of the following TRS is considered.
|
f(a,b) |
→ |
f(a,c) |
(1) |
|
f(c,d) |
→ |
f(b,d) |
(2) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Uncurrying
We uncurry the binary symbol
f
in combination with the following symbol map which also determines the applicative arities of these symbols.
| a |
is mapped to |
a, |
a1(x1) |
| b |
is mapped to |
b, |
b1(x1) |
| c |
is mapped to |
c, |
c1(x1) |
| d |
is mapped to |
d |
There are no uncurry rules.
No rules have to be added for the eta-expansion.
Uncurrying the rules and adding the uncurrying rules yields the new set of rules
|
a1(b) |
→ |
a1(c) |
(6) |
|
c1(d) |
→ |
b1(d) |
(7) |
|
f(a,y1) |
→ |
a1(y1) |
(3) |
|
f(b,y1) |
→ |
b1(y1) |
(4) |
|
f(c,y1) |
→ |
c1(y1) |
(5) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(b) |
= |
6 |
|
weight(b) |
= |
1 |
|
|
|
| prec(c) |
= |
5 |
|
weight(c) |
= |
1 |
|
|
|
| prec(d) |
= |
1 |
|
weight(d) |
= |
1 |
|
|
|
| prec(a) |
= |
7 |
|
weight(a) |
= |
1 |
|
|
|
| prec(a1) |
= |
0 |
|
weight(a1) |
= |
1 |
|
|
|
| prec(c1) |
= |
3 |
|
weight(c1) |
= |
1 |
|
|
|
| prec(b1) |
= |
2 |
|
weight(b1) |
= |
1 |
|
|
|
| prec(f) |
= |
4 |
|
weight(f) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
a1(b) |
→ |
a1(c) |
(6) |
|
c1(d) |
→ |
b1(d) |
(7) |
|
f(a,y1) |
→ |
a1(y1) |
(3) |
|
f(b,y1) |
→ |
b1(y1) |
(4) |
|
f(c,y1) |
→ |
c1(y1) |
(5) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.