3 ms·
Those parentheses are meaningless, since IFF (aka NOT-XOR) is associative and commutative. IFF is a parity counter like XOR. XOR counts how many bits are true
by jolfdb 7y ago
Those parentheses are meaningless, since IFF (aka NOT-XOR) is associative and commutative.
IFF is a parity counter like XOR.
XOR counts how many bits are true. IFF is NOT of how many bits are false. (This shows why XOR is used much more than IFF, because it is cleaner.)
(a XOR false is a, and a XOR true is NOT a, aka a+1 mod 2; a IFF true is a, and a IFF false is a+1 mod 2)
- rrobukef 7y agoThe discussion is the meaninglessness of the parentheses. ⇔ is indeed commutative and associative which means that we can define removing the parentheses as equivalent. However it is equally valid to not do this and define an n-ary '⇔' operator that evaluates ⇔(a1..an) as all ai have the same value. The result is different semantics for A⇔B⇔C. I've taught and always assume ((A⇔B)⇔C). I've apparently used (A⇔B)&(B⇔C) once.