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Korvikt [17]
3 years ago
8

Prove that P (P) = (QA ~ Q)] is a tautology.

Mathematics
1 answer:
alekssr [168]3 years ago
8 0

Answer:

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

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URGENT PLEASE HELP WILL GIVE BRAINLIEST TO CORRECT ANSWER ABC factory brought a pile of coal. If the factory burns 1500 kilogram
wolverine [178]

Solving a system of equations, we will see that there are 6000 kg of coal on the pile.

<h3>How many kilograms of coal are in the pile of coal?</h3>

Let's say that there are N kilograms of coal, and it is planned to burn it totally in D days, these are our variables.

We know that:

  • If the factory burns 1500 kilograms per day, it will finish burning the coal one day ahead of planned.
  • If it burns 1000 kilograms per day, it will take an additional day than planned.

Then we can write the system of equations:

(1500)*(D - 1) = N

(1000)*(D + 1) = N

Because N is already isolated in both sides, we can write this as:

(1500)*(D - 1) = N = (1000)*(D + 1)

Then we can solve for D:

(1500)*(D - 1) =  (1000)*(D + 1)

1500*D - 1500 = 1000*D + 1000

500*D = 2500

D = 2500/500 = 5

Now that we know the value of D, we can find N by replacing it in one of the two equations:

(1500)*(D - 1) = N

(1500)*(5 - 1) = N

(1500)*4= N = 6000

This means that there are 6000 kilograms of coal on the pile.

If you want to learn more about systems of equations:

brainly.com/question/13729904

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