Math Problem Statement
hw14.3
Construct a proof for the argument: P ↔ ¬Q ∴ ¬(P ↔ Q)
1 P ↔ ¬Q
hw14.4
Construct a proof for the argument: ¬(P ↔ Q) ∴ P → ¬Q
1 ¬(P ↔ Q)
hw14.6
Construct a proof for the argument: Fb ∨ ∀xGx ∴ ∀x(Fb ∨ Gx)
1 Fb ∨ ∀xGx
Using the checker:
Rule names (full and abbreviated)
modus ponens ->E
modus tollens MT
modus tollendo ponens DS
double negation DNE
addition vI
adjunction ^I
simplification ^E
bicondition <->I
equivalence <->E
repeat Rep
conditional derivation ->I
reductio ad absurdum RAA
universal instantiation AE
universal derivation AI
existential instantiation EE
existential generalization EI
identity introduction =I
substitution of identicals =E
Solution
Let's break down the problems one by one and construct formal proofs for each.
Problem 14.3
Argument:
∴
Goal: Construct a proof for .
Proof for 14.3:
- (Premise)
- assumes that and have the same truth value (Assumption for reductio ad absurdum).
- By the biconditional elimination on line 2, we have and .
- From (line 1), we also have and .
- Using modus ponens on lines 2 and 4, contradicts , leading to a contradiction.
- Since assuming leads to a contradiction, we conclude . (RAA)
Thus, the argument is valid.
Problem 14.4
Argument:
∴
Goal: Construct a proof for .
Proof for 14.4:
- (Premise)
- Assume (Assumption for conditional derivation).
- From line 1, is false, meaning and have opposite truth values.
- If is true, then must be false, i.e., .
- Therefore, by conditional derivation .
Thus, the argument is valid.
Problem 14.6
Argument:
∴
Goal: Construct a proof for .
Proof for 14.6:
- (Premise)
- To derive , assume is arbitrary.
- Since holds for a particular , it follows that holds for any .
- From line 3, holds for arbitrary .
- By universal derivation (AI), .
Thus, the argument is valid.
These proofs use standard logical rules, such as biconditional elimination, reductio ad absurdum, conditional derivation, and universal instantiation.
Questions:
- How does reductio ad absurdum work in proofs like 14.3?
- Can the biconditional elimination rule always lead to a contradiction?
- Why is universal derivation key in proving arguments like 14.6?
- How can modus tollens help simplify contradictions in proofs?
- What is the difference between conditional derivation and modus ponens?
Tip:
When working with biconditionals, remember they imply two conditional statements in both directions. Always check both implications for consistency.
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Math Problem Analysis
Mathematical Concepts
Propositional Logic
Biconditional Elimination
Reductio Ad Absurdum
Universal Quantification
Formulas
Biconditional: P ↔ Q ↔ (P → Q) ∧ (Q → P)
Modus Ponens: If P → Q and P, then Q
Modus Tollens: If P → Q and ¬Q, then ¬P
Universal Generalization: ∀x(P(x)) if P(x) holds for arbitrary x
Theorems
Biconditional Elimination
Conditional Derivation
Universal Derivation
Suitable Grade Level
Undergraduate
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