discrete math quantifiers

[kilo9]

sorry if that is not the right forum but i did not found discrete math forum

the question:
Hypotheses: everyone in the discrete mathematics class loves proof. someone in the discrete mathematics class has never taken calculus.
conclusion: someone who loves proofs has never taken calculus.

i need to show that the conclusion follows the hypothesis

how i tried to solve it:
p(x) - x is discrete math course
q(x) - x loves proofs
r(x) - x taken calculus
domain of discourse - all people
! = negation

(1) (for all x)(p(x) -> q(x))
(2) (exists x)(p(x) ^ !(r(x))
----------------------------------------
(3) (exists x)(q(x) ^ !r(x))

statement (2):
used conditional, (exists x)!(px(x) -> r(x))
de morgan, !(all x)(p(x) -> r(x))

statement (3): (exist x)(q(x) -> r(x)) (by use of: !(P -> Q) same as P ^ !Q)

after all that a got:
(1) (all x)(p(x) -> q(x))
(2) !(all x)(p(x) -> r(x))
---------------------------
(3) (exists x)(q(x) > r(x))

what i do from here?

 

[HallsofIvy]

From (1) you can replace p(x) in (2) with q(x).

 

[pka]

the question: Hypotheses: everyone in the discrete mathematics class loves proof. someone in the discrete mathematics class has never taken calculus. conclusion: someone who loves proofs has never taken calculus.
i need to show that the conclusion follows the hypothesis how i tried to solve it:
p(x) - x is discrete math course, q(x) - x loves proofs, r(x) - x taken calculus domain of discourse - all people.
after all that a got:
$1.\;\left( {\forall x} \right)\left[ {P(x) \Rightarrow Q(x)} \right]$
$2.\;\left( {\exists x} \right)\left[ {Q(x) \wedge \neg R(x)} \right]/\therefore \left( {\exists x} \right)\left[ {Q(x) \wedge \neg R(x)} \right]$

As you can see I have changed the notation to standard form.
Having taught discrete mathematics classes over some forty years, I have ever seen a textbook that requires students to use $\bf{EI}$ Existential Instantiation ,$\bf{EG}$ Existential Generalization, or $\bf{UI}$ Universal Instantiation. But you need them for this proof.
$1.\;\left( {\forall x} \right)\left[ {P(x) \Rightarrow Q(x)} \right]$
$2.\;\left( {\exists x} \right)\left[ {P(x) \wedge \neg R(x)} \right]/\therefore \left( {\exists x} \right)\left[ {P(x) \wedge \neg R(x)} \right]$
$3.\;P(w)\wedge \neg R(w)$ 2, $\bf{EI}$
$4.\;P (w) \Rightarrow Q(w)$, 1. $\bf{UI}$
$5,\;P(w)$ 3. Simp.
$6.\:Q(w)$ 4. & 5. M.P.
$7.\;\neg R(w)$ 3. Simp.
$8.\; Q(w)\wedge \neg R(w)$ 5. & 7. Conjunction.
$9.\; \left( {\exists x} \right)\left[ {Q(x) \wedge \neg R(x)} \right]$ 8, $\bf{EG}$