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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.

edited | 927 views
0
how can we solve this question using Rule of inference .

$A\wedge(A\to (B\vee C))\wedge ( B\to \neg A)$
$\equiv A \wedge (\neg A \vee B \vee C)(\neg A \vee \neg B)$
$\equiv (A\wedge \neg B) \wedge (\neg A \vee B \vee C)$
$\equiv (A\wedge \neg B \wedge C)$
$C$ 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 (61k points)
edited by
$A$ $B$ $C$ $A\to \left(B \vee C\right)$ $B\to\neg A$ $A\wedge \left(A \to \left(B \vee C\right)\right) \wedge \left( B \to \neg A\right)$ $C$
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 (5.8k points)
+1

Anu

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

0

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

$C$ 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

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. $(\neg B \wedge (B \vee C)) \rightarrow C$

by Junior (985 points)