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+6 votes
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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.
asked in Mathematical Logic by Veteran (59.5k points)
edited by | 589 views
0
how can we solve this question using Rule of inference .

3 Answers

+17 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 $A\wedge (A\to (B\vee C))\wedge (B\to \neg A).$
answered by Veteran (55.4k points)
edited by
+6 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.
answered by Loyal (5.9k points)
0

Anu

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

+5 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 $

answered by Junior (983 points)


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