logic basic

Question

Let P(x) , Q(x) and R(x) be the statements "x is a clear explantion " , "x is satisfactory " and "x is an excuse " resp

Suppose that the domain x consists of all English text

Express each of these using quantifiers , logical connectives and p(x) R(x) and q(x)

1) All clear explanation are satisfactory

2) Some excuses are unsatisfactory

3)  Some excuses are not clear explanation

4) does 3 follow from 1and 2

Answer 1

Please just point out my mistakes

1)∀x(P(X)--->Q(x))

2)∃x(Q(x)∧ ∼R(x))

3)∃x(Q(x)∧ ∼P(x)) D)

I dont think it follows from the first statement all clear statemet are satisfactory ---1

it doesnt imply all the satisfactory thing are just clear statements only . ------------------2

Some excuses are unsatisafctory

it can be said that some excuses are satisfactory too

but i cant say some excuses are not clear explanation (from 2)

Comments

your expression for all statements are correct ,,

but 3 rd one  follow by 1, 2 .. i solved it by logical deduction .. but we can also proof by inference . and logical deductions ( using these                rules http://aptitude.gateoverflow.in/2346/cat1999-125#a2370) work here also . i guess .

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okay thanks :) ya i will see that thank you

Answer 2

Solving for d)

 

$\forall x \big( P(x) \rightarrow Q(x) \big)$---------------(1)

$\text{can be re-written as}$

 $\bigg(P_{1}(x) \rightarrow Q_{1}(x)\bigg) \color{red}\wedge \bigg(P_{2}(x) \rightarrow Q_{2}(x)\bigg) \color{red}\wedge … \color{red} \wedge \bigg(P_{n}(x) \rightarrow Q_{n}(x)\bigg)$

$\text{which further can be written as}$

$\bigg(\overline{P_{1}(x)}  \space  \vee Q_{1}(x)\bigg)  \space \color{Blue} \wedge \bigg(\overline{P_{2}(x)}  \space  \vee Q_{2}(x)\bigg)  \space \color{Blue} \wedge \space… \space \color{Blue}\wedge \bigg(\overline{P_{n}(x)}  \space  \vee Q_{n}(x)\bigg) $ -----------------(2)

 

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$\exists x \bigg( R(x) \wedge \space Q(x) \bigg)$ --------------(3)

$\text{can be re-written as}$

$\bigg(R_{1}(x) \wedge \overline{Q_{1}(x)} \bigg) \vee \bigg(R_{2}(x) \wedge \overline{Q_{2}(x)} \bigg) \vee \space … \vee \space \bigg(R_{n}(x) \wedge \overline{Q_{n}(x)} \bigg)$----------(4)

 

$\text{They are asking whether c can be inferred from premises 2 and 4 ?}$

$\text{Let us check}$

 

$\bigg(\overline{P_{1}(x)}  \space  \vee Q_{1}(x)\bigg)  \space \color{Blue} \wedge \bigg(\overline{P_{2}(x)}  \space  \vee Q_{2}(x)\bigg)  \space \color{Blue} \wedge \space… \space \color{Blue}\wedge \bigg(\overline{P_{n}(x)}  \space  \vee Q_{n}(x)\bigg) $ $\color{purple}\wedge$

$\bigg(R_{1}(x) \wedge \overline{Q_{1}(x)} \bigg) \vee \bigg(R_{2}(x) \wedge \overline{Q_{2}(x)} \bigg) \vee \space … \vee \space \bigg(R_{n}(x) \wedge \overline{Q_{n}(x)} \bigg)$

 

If you solve it further, you will get

 

$\forall x \bigg( R(x) \wedge \overline{P(x)} \wedge \overline{Q(x)} \bigg)$

which can be written as 

$\forall x  \bigg( R(x) \wedge \overline{P(x)} \bigg) \wedge \forall x \bigg( \overline{Q(x)} \bigg)$

 

$ \forall \space can \space be \space replaced \space  by \space \exists \space , \space hence \space $

$\exists x  \bigg( R(x) \wedge \overline{P(x)} \bigg) \wedge \exists x \bigg( \overline{Q(x)} \bigg)$

 

$\text{So in my opinion, indeed C can be inferred from both a and b}$