1. (∃x)(Hx • ∼Ix)
2. (∀x)(Hx ⊃ Jx)
3. ∼(∃x)(Jx • Kx)
-Which of the following propositions is derivable from the given premises in M?
A) (∃x)Ix
B) (∀x)∼(Ix $ \lor $ Kx)
C) (∃x) ∼Hx
D) (∀x)∼(Ix • Kx)
E) (∃x)∼(Ix $ \lor $ Kx)

Question

Quizplus · Accepted Answer

From premise 1, we can derive:
- (∃x)Hx (by existential instantiation)
From premise 2, we can derive:
- (∀x)Jx ∨ ∼Hx (by contrapositive)
From premise 3, we can derive:
- ∼(∃x)Jx ∨ ∼(∃x)Kx (by De Morgan's and quantifier negation)
- (∀x)∼Jx ∨ (∀x)∼Kx (by quantifier negation)

To derive (∃x)∼(Ix $ \lor $ Kx), we can use proof by contradiction. Assume ¬(∃x)∼(Ix $ \lor $ Kx), i.e., (∀x)(Ix $ \lor $ Kx). Then, we can derive:
- (∀x)∼∼(Ix ⊃ Kx) (by contrapositive)
- (∀x)(Ix ⊃ Kx) (by double negation)
- (∀x)(∼Kx ⊃ ∼Ix) (by contrapositive)
- (∀x)(Jx ⊃ ∼Ix) (by premise 2 and transitivity)
- (∀x)∼Jx (by proof by contradiction and disjunctive syllogism with premise 3)
- (∀x)∼(Hx ⊃ Jx) (by contrapositive)
- (∀x)(∼Jx ⊃ ∼Hx) (by contrapositive)
- (∀x)∃y∼Hy (by premise 1 and universal instantiation)
- (∀x)(∃y∼Hy ⊃ ∃y∼Ix) (by universal instantiation)
- (∀x)∃y∼Ix (by reasoning by cases, since either ∼Hy or ∃y∼Ix must hold for each x)
- ∃x∃y∼Ix (by existential instantiation)
- (∃x)∼(Ix ⊃ Kx) ∧ (∃x)∃y∼Ix (by conjunction)
- (∃x)∼((Ix ⊃ Kx) ⊃ (∀y)∼Iy) (by existential generalization)
- (∃x)∼((∀y)(Ix ⊃ Kx) ⊃ (∀y)∼Iy) (by universal instantiation)
- (∃x)∼(∀y)((Ix ⊃ Kx) ⊃ ∼Iy) (by quantifier exchange)
- (∃x)(∃y)((Ix ⊃ Kx) ∧ Iy) (by De Morgan's and quantifier negation)
- (∃x)∃y(Ix ∧ Iy ∧ Kx) (by reasoning by cases, since either Iy or Kx must hold for each x and y)
- (∃x)Ix ∧ (∃y)(Iy ∧ Kx) (by existential generalization)
- (∃x)Ix (by existential instantiation)

This leads to a contradiction with our assumption. Therefore, we can conclude (∃x)∼(Ix $ \lor $ Kx).

1 (∃X)(Hx • ∼Ix) 2 ∨ \lor ∨ Kx) C) (∃X) ∼Hx D) (∀X)∼(Ix • Kx) E) (∃X)∼(Ix

1 (∃X)(Hx • ∼Ix)
2 $\lor$ Kx)
C) (∃X) ∼Hx
D) (∀X)∼(Ix • Kx)
E) (∃X)∼(Ix