Solution to Use inference rules to deduce the following conclusions from the following sets of premises: a) … - Sikademy

Sept. 17, 2019

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Archangel Macsika

Use inference rules to deduce the following conclusions from the following sets of premises: a) Premises: p ∨ q q → r, p ∧ s → t, ¬r, ¬q → u ∧ s Conclusion: t b) Premises: (p ∧ t) → (r ∨ s), q → (u ∧ t), u → p, ¬s Conclusion: q → r

a. Solution for - Premises: p ∨ q, q → r, p ∧ s → t, ¬r, ¬q → u ∧ s Conclusion: t

1q → rPremise
2¬ rPremise
3¬ qModus Tollens of 1 and 2
4p ∨ qPremise
5pDisjunctive syllogism of 5 and 4
6¬q → u ∧ sPremise
7¬ qCopy 3
8u ∧ sModus Ponens of 7 and 6
9sSimplification of 8
10pCopy 5
11p ∧ sConjunction of 9 and 10
12p ∧ s → tPremise
13tModus Ponens of 11 and 12

By arriving at the conclusion "t", the argument is proven.

∴ The premise and conclusion is valid.

b. Solution for - Premises: (p ∧ t) → (r ∨ s), q → (u ∧ t), u → p, ¬s Conclusion: q → r

Since the above argument includes a proposition "q", the argument above will be valid, if the argument with the following premise and conclusion below is valid.

Knowing this, we can rewrite
Premises: (p ∧ t) → (r ∨ s)
q → (u ∧ t)
u → p
¬s
Conclusion: q → r
as
Premise: (p ∧ t) → (r ∨ s)
q → (u ∧ t)
u → p
¬s
q
Conclusion: r

Observe the two propositions and how they differ from each other. If we can prove that the second proposition is valid, it automatically means that the first (original) proposition is also valid.

1q → (u ∧ t)Premise
2q → uSimplification of 1
3u → pPremise
4q → pHypothetical Syllogism of 2 and 3
5q → tSimplication of 1
6q → (p ∧ t)Conjunction of 4 and 5
7(p ∧ t) → (r ∨ s)Premise
8q → (r ∨ s)Hypothetical Syllogism of 6 and 7
9qPremise
10r ∨ sModus Ponens of 8 and 9
11¬sPremise
12rDisjunctive syllogism of 10 and 11

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