Quick predicate logic and quantifier homework check (2)

Question

I need a quick check on my homework to see if any questions are wrong. We just finished learning quantifiers (for all, for every) yesterday and the assignment is due today at 4pm MST. If it is wrong, please point me in the right direction or the answer/why.

My answers:

(a) Ch → Ce

(b) L(e,a) ∨ ¬ L(a,e)

(c) [(L(m,a) ∧ Ca) → L(m,...) NOT DONE

(d) ∀x(Cx → L(x,m))

(e) ∀x(Cx → L(m,x) ∧ ∃x(¬Cx → L(a,x))

(f) ∀(Cx → ¬Rx) ∧ &all;y(Ry → ¬Cy)

(g) ∃x[(Cx ∧ Ry) ∧ L(x,y)]

(h) ∀x∃y[((Rx ∧ Cy) → L(x,y)) ∧ ¬L(y,a)]

(i) ∃x∀yL(x,y) → ∀y∃xL(y,x)

(j) (L(m,a) ∧ Ce) → [Rm →∃x∃yL(x,y)]

(k) ∃x[(Cx ∧ L(x,m)) → L(e,m)]

(i) L(m,h) ↔ ∃x(Rx ∧ L(x,e))

Answer 1

(c) Requires an existential quantifier on the part that is not done.

(d) You have the implication the wrong way round. You have written "all Celts like Marcus". "Only Celts like Marcus" means that if x likes Marcus, x is a Celt.

(e) Two errors here: the first implication is the wrong way round, as in (d), and the second implication should be a conjunction.

(g) This is missing a quantifier. The variable y needs to be universally quantified.

(h) Two errors here: in the first conjunct, the Cy should be on the right of the implication, not the left, otherwise the implication is trivially true in the event that there exists at least one non-Celt. The second conjunct has the wrong variable in it.

(i) Come on, you're not trying here: where are the predicates for Roman and Celt?

(j) Looks like a typo: you have 'a' instead of 'e'.