Update:
i can use these :
1) d->(y->d)
2) (d->(y->χ))->((d->y)->(d->χ))
3) (查-> 流)->((查->y)->d)
and Modus Ponens..
any1 ? pls
Hello ppl,
i am a bit struggling here, i ve been asked to to prove these (below) without using any known theorems , how am i supposed to do that ? any1 knows? Help!
(a) d |− 流 → d
(b) {d, 查}|− y (you can use (a)).
(c) 洵d |- ((查 → 查) → d)
my nerves, ah .
(a) is pretty simple; the length of the shortest derivation is 3.
(b): from (a), you have 查 |− 流 → 查 and d |− 流 → d. Now use axiom 3 (where d and y are switched).
(c): again, the formula right of |- seems like the end of axiom 3 where y is 查. This means the beginning of this instance of axiom 3 is 查 -> 洵d. This is derivable from 洵d by (a).
For (a), there exists a derivation consisting of 3 formulas. One of them has to be the assumption d. The second has to be an axiom. Then the third is obtained from the first two by MP.
For the rest, please describe what exactly you don't understand. For example, a hint for (b) says, "from (a), you have 查 |− 流 → 查". Surely you understand that if you take a derivation of d |− 流 → d from (a) and replace d with 查, you get a derivation of 查 |− 流 → 查.
As a general remark, I would not try solving this problem without first studying several examples of derivations from the textbook or lecture notes. You need to understand how this calculus works and have some intuition about it.