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Determine whether the following arguments are either valid or invalid by using the indirect method only of establishing validity. Circle your answer. Show all of your work for full credit. 1. 1. G⊃(I∨D)/(I⋅D)⊃B//∼G⊃B 2. (∼J∙∼K)/(L⊃J)/(M⊃K)/(M⊃∼L)⊃∼(N∙O)//∼N 3. ∼(O⋅Z)⊃(M∼A)/M⊃R/Z≡∼O/∼R∨A//∼O≡∼R (Z⋅K)v∼(R⊃O)/(O∨M)⊃∼R/(M⋅K)≡R//∼Z≡O (A⊃B)⊃(C⋅D)/(∼A∨∼B)⊃E/∼E∥(∼C⋅∼D)⊃∼E B⊃(E∙D)/(∼Ev∼F)/E⊃(B∨G)/G⊃(D⊃F)//Gv∼E

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To determine the validity of each argument using the indirect method, we will assume the negation of the conclusion and try to derive a contradiction. If we can derive a contradiction, then the original argument is valid. If not, the argument is invalid.

G⊃(I∨D)

I⋅D⊃B

∼G⊃B

Assume ∼(∼G⊃B) (negation of the conclusion): G∧∼B

G (Assumption)

G⊃(I∨D) (Premise 1)

I∨D (Modus Ponens 1, 2)

I∨D⊃B (Premise 2)

B (Modus Ponens 3, 4)

∼B (Simplification 5, 2nd conjunct)

B∧∼B (Conjunction 5, 6)

∼G (Reductio ad absurdum 1-7)

G∧∼G (Conjunction 1, 8)

Since we derived a contradiction, the assumption ∼(∼G⊃B) leads to an inconsistency. Therefore, the argument is valid. The conclusion ∼G⊃B holds.

(∼J∙∼K)

L⊃J

M⊃K

M⊃∼L

∼(N∙O)

∼N

Assume ∼∼N (negation of the conclusion): N

(∼J∙∼K) (Premise 1)

L⊃J (Premise 2)

M⊃K (Premise 3)

M⊃∼L (Premise 4)

∼(N∙O) (Premise 5)

N (Assumption)

N∙O (Conjunction 6, 5)

∼(N∙O) (Premise 5)

N∙O∧∼(N∙O) (Conjunction 7, 8)

∼N (Reductio ad absurdum 6-9)

N∧∼N (Conjunction 6, 10)

Since we derived a contradiction, the assumption ∼∼N (N) leads to an inconsistency. Therefore, the argument is valid. The conclusion ∼N holds.

∼(O⋅Z)⊃(M∼A)

M⊃R

Z≡∼O

∼O⊃∼R∨A

∼O≡∼R

Assume ∼(∼O≡∼R) (negation of the conclusion): ∼O∧R

∼(O⋅Z)⊃(M∼A) (Premise 1)

M⊃R (Premise 2)

Z≡∼O (Premise 3)

∼O⊃∼R∨A (Premise 4)

∼O≡∼R (Assumption)

∼O∧R (Assumption)

∼O (Simplification 6)

∼R

User Corinne Kubler
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