Prove that the conjunction (A --> B) and (B -->C) and not(A --> C) is false usingpropositional logic

Name: Prove that the conjunction (A --> B) and (B -->C) and not(A --> C) is
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Description: Prove that the conjunction (A --> B) and (B -->C) and not(A --> C) is false usingpropositional logic

Given:(A→B) and (B →C) and not (A→C) First Change this statement into symbolic form:…

Answered: Prove that the conjunction (A --> B)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Prove that the conjunction (A --> B) and (B --
>C) and not(A --> C) is false using
propositional logic

Expert Solution

Step 1

$G i v e n : (A \to B) a n d (B \to C) a n d n o t (A \to C)$

First Change this statement into symbolic form:

$(A \to B) \land (B \to C) \land ~ (A \to C)$ ....(1)

We know that By Conditional Syllogism

$(A \to B) \leftrightarrow (~ A \lor B)$

Use this in equation (1)

$(~ A \lor B) \land (~ B \lor C) \land ~ (~ A \lor C)$

Step 2

By using De-Morgan's law and Double negation

$(~ A \lor B) \land (~ B \lor C) \land (A \land ~ C)$

By using Distributive law, we will get

$[(~ A \lor B) \land ~ B) \lor ((~ A \lor B) \land C)] \land (A \land ~ C)$

$[(~ A \land ~ B) \lor (B \land ~ B)) \lor ((~ A \land C) \lor (B \land C)] \land (A \land ~ C)$

$[((~ A \land ~ B) \lor F) \land (A \land ~ C)] \lor [(~ A \land C) \land (A \land ~ C)) \lor ((B \land C) \land (A \land ~ C)]$

$[(~ A \land ~ B) \land (A \land ~ C)] \lor [(~ A \land C) \land A) \land (~ A \land C) \land ~ C)) \lor ((B \land C) \land A) \land (B \land C \land ~ C)]$

$[(~ A \land ~ B) \land A \land (~ A \land ~ B) \land ~ C] \lor [(~ A \land A) \land (C \land A)) \land ((~ A \land ~ C) \land (C \land ~ C)) \lor ((B \land A) \land (C \land A)) \land ((B \land ~ C) \land (C \land ~ C)]$

Since we know that By Negation Law $p \land ~ p \equiv c$ where p is any statement and c is contradiction means false, we are using F here in place of c.

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