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14 votes
14 votes
Show that proposition $C$ is a logical consequence of the formula$$A\wedge \left(A \to \left(B \vee C\right)\right) \wedge \left( B \to \neg A\right)$$using truth tables.
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how can we solve this question using Rule of inference .
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3 Answers

29 votes
29 votes
Best answer
$A\wedge(A\to (B\vee C))\wedge ( B\to \neg A)$
$\equiv A \wedge (\neg A \vee B \vee C)(\neg A \vee \neg B)$
$\equiv (A\wedge \neg B) \wedge (\neg A \vee B \vee C)$
$\equiv (A\wedge \neg B \wedge C)$
$C$ is logical consequence of a formula $X$ if, $X\to C$ is true.

$\begin{align}\text{Here, }X &\equiv A\wedge (A\to (B\vee C))\wedge ( B\to \neg A)\\ &\equiv A\wedge \neg B \wedge C\end{align}$

Checking,   
$(A\wedge \neg B \wedge C)\to C$
$\equiv  \neg(A\wedge \neg B \wedge C) \vee C$
$\equiv \neg A \vee B \vee \neg C \vee C$
$\equiv 1.$

So, $C$ is logical consequence of $A\wedge (A\to (B\vee C))\wedge (B\to \neg A).$
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2 Comments

≡(A∧¬B)∧(¬A∨B∨C)≡(A∧¬B)∧(¬A∨B∨C)
≡(A∧¬B∧C) (How to get this from above ?)
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This is how we can get,

(A∧¬B)∧(¬A∨B∨C)

≡ (A∧¬B∧¬A) ∨ (A∧¬B∧B) ∨ (A∧¬B∧C)

≡ F ∨ F ∨ (A∧¬B∧C)

≡ (A∧¬B∧C)
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10 votes
10 votes
$A$ $B$ $C$ $A\to \left(B \vee C\right)$ $B\to\neg A$ $A\wedge \left(A \to \left(B \vee C\right)\right) \wedge \left( B \to \neg A\right)$ $C$
T T T T F F T
T T F T F F T
T F T T T T T
T F F F T F T
F T T T T F T
F T F T T F T
F F T T T F T
F F F T T F T
  1. Logical consequence (also entailment) is one of the most fundamental concepts in logic. It is the relationship between statements that holds true when one logically "follows from" one or more others.
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3 Comments

Anu

In the last column how to write all C is true??

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@Lakshman Patel RJIT @Anu @srestha
 

How this last column entries are all true? Is it because X = A ∧ (A → (B ∨ C)) ∧ (B → ¬A) and X → C = T.
Can someone clear this doubt please?

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edited by

@vupadhayayx86

$C$ is a logical consequence of a formula $X$ if, $X→C$ is true.

$A$ $B$ $C$ $A\to \left(B \vee C\right)$ $B\to\neg A$ $A\wedge \left(A \to \left(B \vee C\right)\right) \wedge \left( B \to \neg A\right)$ $\bigg[A\wedge \left(A \to \left(B \vee C\right)\right) \wedge \left( B \to \neg A\right)\bigg] \rightarrow C$
T T T T F F T
T T F T F F T
T F T T T T T
T F F F T F T
F T T T T F T
F T F T T F T
F F T T T F T
F F F T T F T

 

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7 votes
7 votes

Using Inference Rules,

Modus Ponens, 
1. $(A \wedge (A \rightarrow (B \vee C ))) \rightarrow (B \vee C)$

2. $(B \rightarrow \neg A) \equiv (\neg B \vee \neg A)$

Modus Tollens,
3. $((\neg B \vee \neg A) \wedge A) \rightarrow \neg B$

Disjunctive syllogism,
4. $(\neg B \wedge (B \vee C)) \rightarrow C $

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