how can we solve this question using Rule of inference .

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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 $A\wedge (A\to (B\vee C))\wedge (B\to \neg A).$

$\equiv A \wedge (\neg A \vee B \vee C)(\neg A \vee \neg B)$

$\equiv (A\wedge \neg B) \wedge (\neg A \vee B \vee C)$

$\equiv (A\wedge \neg B \wedge C)$

$C$ 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 $A\wedge (A\to (B\vee C))\wedge (B\to \neg A).$

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 |

@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?

0

edited
Oct 8, 2019
by Lakshman Patel RJIT

$C$ 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 |

4

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 $