# Quick predicate logic and quantifier homework check

I need a quick check on my homework to see if any questions are wrong. We just finished learning quantifiers (for all, for every) today and the assignment is due tomorrow.

(a) Lr ∧ Lb
(b) ¬Dr
(c) Db ↔ Dm
(d) Lc ∧ ¬Dc
(e) Bvk ∧ Bak
(f) ∃x(Lx ∧ ¬Dx)
(g) (La ∧ Da) → (Lc ∧ ¬Dc)
(h) Bck

• I may not be an expert in logics, but looks just fine. Nov 9, 2016 at 5:15
• g and h are wrong. The error in h is insignificant. h should (Dc v ~Dc) -> Bck. g is backwards -- "only if" makes what follows "only if" the antecedent. Nov 9, 2016 at 5:42
• @virmaior h) My Professor said that "whether or not" is a dummy operative, and there was an example he gave that didn't have the "whether or not" symbolized. g) My Professor said today that "only if" governs the consequent. Is that equivalent to what you said? Nov 9, 2016 at 5:50
• @virmaior: As I had to look it up myself - 'p only if q' means that if p happened, q is implied, because without it, p wouldn't have happened (necessary condition). Whereas 'p if q' exactly means q -> p, because it means "if q, then p". Therefore, I'd argue that G is correct. Regarding (h) it comes down to conventions I guess. Writing out tautologies and crossing them out afterwards may help showing that you understood what you're doing, but if the teacher does not like it, it's better not to do it. Nov 9, 2016 at 6:34
• Regarding the "dummy operator", it makes sense. If we introduce (as suggested) the "dummy" antecedent (Dc v ~Dc), due to the fact that it is always TRUE, the truth-value of the conditional (Dc v ~Dc) → Bck will be the same of Bck. Nov 9, 2016 at 7:25