7a=7(ava)=a↓a
a^b=7(7(a^b))=7(7av7b)=7a↓7b=(a↓a)↓(b↓b)
avb=7(7(avb))=7(7a^7b)=7a|7b=(a↓a)|(b↓b)
I used DeMorgan's laws up there btw. and the fact that double negatives cancel.
Marc
How do I completely solve this problem? I have created truth tables for a↓b and a|b but I don't understand the rest of the question.
Here it is:
Connectives ↓ (Pierces's arrow, nor) and | (Scheffer's stroke, nand) are defined by a↓b := 7(a v b), and a|b := 7(a ^b). Express each of these connectives in terms of 0 and →. Construct truth tables for ↓ and |. Show that each of the connectives 7, ^ and v can be expressed in terms of the connective ↓, and in terms of the connective |.
*the 7 is to negate....