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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.
in Mathematical Logic by Veteran (52.2k points)
edited by | 861 views
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how can we solve this question using Rule of inference .

3 Answers

+21 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 $A\wedge (A\to (B\vee C))\wedge (B\to \neg A).$
by Veteran (60.5k points)
edited by
+7 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.
by Loyal (5.7k points)
+1

Anu

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

0

@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?

+1

@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

 

+6 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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