Mathematics, 03.10.2019 02:20 jaydostm9540

Prove that p (p) = (qa ~ q)] is a tautology.

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Answer from: kris22elizondop9v1bb

The statement $P \leftrightarrow [(\lnot P) \rightarrow (Q \land \lnot Q)]$ is a tautology.

Step-by-step explanation:

A tautology is a formula which is "always true" that is, it is true for every assignment of truth values to its simple components.

To show that this statement is a tautology we are going to use a table of logical equivalences:

$P \leftrightarrow [(\lnot P) \rightarrow (Q \land \lnot Q)] \equiv$

$\equiv (P \land [(\lnot P)\rightarrow (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [(\lnot P)\rightarrow (Q \land \lnot Q)])$ by the logical equivalences involving bi-conditional statements

$\equiv (P \land [\lnot(\lnot P)\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [\lnot(\lnot P)\lor (Q \land \lnot Q)])$ by the logical equivalences involving conditional statements

$\equiv (P \land [P\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [ P\lor (Q \land \lnot Q)])$ by the Double negation law

$\equiv (P \land [P\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot P\land \lnot(Q \land \lnot Q))$ by De Morgan's law

$\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot(Q \land \lnot Q))$ by the Negation law

$\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot Q \lor \lnot(\lnot Q))$ by De Morgan's law

$\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot Q \lor Q)$ by the Double negation law

$\equiv (P \land P) \lor(\lnot P \land \lnot P\land \lnot Q \lor Q)$ by the Identity law

$\equiv (P) \lor(\lnot P \land \lnot P\land \lnot Q \lor Q)$ by the Idempotent law

$\equiv (P) \lor(\lnot P \land \lnot P\land (Q\lor \lnot Q))$ by the Commutative law

$\equiv (P) \lor(\lnot P \land \lnot P\land T)$ by the Negation law

$\equiv (P) \lor(\lnot (P \lor P)\land T)$ by De Morgan's law

$\equiv (P) \lor(\lnot (P)\land T)$ by the Idempotent law

$\equiv (P \lor\lnot P) \land(P \lor T)$ by the Distributive law

$\equiv (T) \land(P \lor T)$ by the Negation law

$\equiv (T) \land(T)$ by the Domination law

$\equiv T$

Prove that p (p) = (qa ~ q)] is a tautology.

Answer from: Quest

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the (original price) is $420.

0.7x = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

Answer from: Quest

alot

step-by-step explanation:

its a wood chuck that can chuck wood

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