Propositional Logic Proofs: Dilemmas and Rules
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Questions and Answers

Provide the missing step and the rule applied for the following proof:

  1. P → Q
  2. R → S
  3. P ∨ R
  4. ?

Q ∨ S, CD (Constructive Dilemma), lines 1, 2, and 3

Provide the missing step and the rule applied for the following proof:

  1. (P → Q) & (R → S)
  2. P ∨ R
  3. ?

Q ∨ S, CCD (Complex Constructive Dilemma), lines 1, and 2

Provide the missing step and the rule applied for the following proof:

  1. P → Q
  2. P → R
  3. ¬Q ∨ ¬R
  4. ?

¬P, DD (Destructive Dilemma), lines 1, 2, and 3

Provide the missing step and the rule applied for the following proof:

  1. (P → Q) & (R → S)
  2. ¬Q ∨ ¬S
  3. ?

<p>¬P ∨ ¬R, DCD (Complex Destructive Dilemma), lines 1, and 2</p> Signup and view all the answers

Complete the following proof using the rules provided and indicate the rule:

  1. A → B
  2. A
  3. ?

<p>B, MP (Modus Ponens), lines 1 and 2</p> Signup and view all the answers

Complete the following proof using the rules provided and indicate the rule:

  1. ¬¬P
  2. ?

<p>P, DN (Double Negation), line 1</p> Signup and view all the answers

Complete the following proof using the rules provided and indicate the rule:

  1. P
  2. Q
  3. ?

<p>P &amp; Q, Conj (Conjunction), lines 1 and 2</p> Signup and view all the answers

Complete the following proof using the rules provided and indicate the rule:

  1. P ∨ Q
  2. ¬P
  3. ?

<p>Q, DS (Disjunctive Syllogism), lines 1 and 2</p> Signup and view all the answers

Flashcards

Complex Constructive Dilemma

If P then Q; If R then S; Either P or R; Therefore, either Q or S

Complex Destructive Dilemma

If P then Q; If P then R; Either not Q or not R; Therefore, not P

Hypothetical Syllogism (HS)

If P then Q; If P then R; Therefore, if P then both Q and R

Simplify (Simp)

From the conjunction of two statements, infer either statement separately

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Double Negative (DN)

If A is true, then not not A is true

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Disjunctive Syllogism (DS)

Given 'A or B' and 'not A', infer B

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Conjunction (Conj)

From 'A' and 'B', infer 'A and B'

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Disjunction (Disj)

From 'A' infer 'A or B'

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De Morgan’s Laws (DM)

Rules to convert between conjunctions/disjunctions and their negations

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Study Notes

  • Proofs will cover Simple Constructive Dilemma, Complex Constructive Dilemma, Simple Destructive Dilemma, and Complex Destructive Dilemma.
  • Permitted rules include Modus Ponens (MP), Modus Tollens (MT), Double Negative (DN), Simplification (Simp), Conjunction (Conj), Conjunctive Syllogism (CS), Disjunctive Syllogism (DS), Disjunction (Disj), De Morgan’s Laws (DM), Conditional Proof (CP), Hypothetical Syllogism (HS), and Biconditional Equivalence (BE).
  • Use the following symbols:
    • ¬A — for ‘it’s not the case that A’
    • A→B — for ‘if A then B’
    • A & B — for ‘A and B’
    • A ∨ B – for ‘A or B’
    • A ↔ B — for ‘A if and only if B’

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Description

Explore propositional logic proofs focusing on Simple and Complex Constructive and Destructive Dilemmas. Learn to apply rules such as Modus Ponens, Modus Tollens, and De Morgan’s Laws. Grasp symbolic logic using negation, conditionals, conjunctions, disjunctions, and biconditionals.

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