# Proof With and Without Truth Tables

$(6)$

Use truth tables to determine whether or not the following argument is correct: "If the tax rate and the unemployment rate both go up, then there will be a recession. If the GNP goes up, then there will not be a recession. The GNP and taxes are both going up. Therefore, the unemployment rate is not going up." In other words, decide whether the concluding statement must be true, given that the preceding compound statements are true.

My Work

Let $P :=$ "The tax rate is going up." Let $Q :=$ "The unemployment rate is going up." Let $R :=$ "There is a recession." Let $S :=$ "The GNP is going up." Then "If the tax rate and the unemployment rate both go up, then there will be a recession" has the logical form $(P \wedge Q) \Longrightarrow R$. Likewise, the statement "If the GNP goes up, then there will not be a recession" has the logical format $S \Longrightarrow \sim R$. Using truth tables, we will prove the truth or falsity of the statement "The GNP and taxes are both going up. Therefore, the unemployment rate is not going up" which can be expressed as follows: $(S \wedge P) \Longrightarrow \sim Q$ which, by a previously given equivalence, is equivalent to $\sim (S \wedge P) \vee \sim Q$ .

$(7)$

Do $(6)$ without truth tables.

Proof: (My Work) From a previous question $(5)$ I have the following equivalences:

Use the following equivalances: $$P \Longrightarrow Q \equiv \ \sim P \vee Q \qquad \sim \sim P \equiv P \qquad \sim (P \wedge Q) \equiv \sim P \ \vee \sim Q$$ to prove that $P \Longrightarrow Q \vee R \equiv P \wedge \sim Q \Longrightarrow R$.

\begin{align*} (S \wedge P) \Longrightarrow \sim Q &\equiv \ \sim (S \wedge P) \vee \sim Q \\ \sim (S \wedge P) \vee \sim Q &\equiv \ \sim(S \wedge P \wedge Q) \\ \end{align*} Since $S \Longrightarrow \sim R$ and $(P \wedge Q) \Longrightarrow R$, $(S \wedge P \wedge Q) \equiv (R \wedge \sim R)$. By the Law of the Excluded Middle, $(R \wedge \sim R)$ is false for all $R$, so $\sim(R \wedge \sim R)$ is true for all $R$, so we have that $(S \wedge P) \Longrightarrow \sim Q$ is a true statement.

My Question

I can do the proof without truth tables, but I am not sure what I should do to prove it with truth tables. Should I construct the truth values of every possible configuration of $P, Q, R, S$ and piece together the following statement:

Therefore, the unemployment rate is not going up" which can be expressed as follows: $(S \wedge P) \Longrightarrow \sim Q$ which, by a previously given equivalence, is equivalent to $\sim (S \wedge P) \vee \sim Q$ .

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Yes, you have to write down all cases of true / false for P,Q,R,S and find the case which fits to your statements (you have 2^4 cases) – ulead86 Aug 30 '12 at 11:06

If you number the statements as

1. If the tax rate and the unemployment rate both go up, then there will be a recession
2. If the GNP goes up, then there will not be a recession
3. The GNP and taxes are both going up

then you can exclude all but one of the rows of the truth table as being contrary to these statements

P Q R S Comment

T T T T No - 2
T T T F No - 3
T T F T No - 1
T T F F No - 1, 3
T F T T No - 2
T F T F No - 3
T F F T
T F F F No - 3
F T T T No - 2, 3
F T T F No - 3
F T F T No - 3
F T F F No - 3
F F T T No - 2, 3
F F T F No - 3
F F F T No - 3
F F F F No - 3


That leaves $P$ and $S$ true and $R$ and $Q$ false. In particular $Q$ is false and so unemployment is not going up.

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You need to construct a table with one line for every possible assignment of T or F to the four propositions P, Q, R, S. There are 16 cases. Then work out on each assignment the truth value of each premiss and of the conclusion. Is there an assignment, a line of the table, where the premisses are all true and conclusion false? If so, then the argument is not tautologically valid. Otherwise, i.e. if every assignment which makes the premisses true makes the conclusion true, the argument is tautologically valid.

This is really basic stuff, and the fact that you are asking about it suggests you should have a look at another textbook (always a very good idea to read more than one presentation of elementary logical ideas). There are dozens out there, including -- ahem, cough -- Peter Smith's Introduction to Formal Logic (CUP) which is jolly clear on such stuff, though I say so myself! :-)

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sorry professor i changed my opinion, actually self publicity is allowed here, can you please edit your post so that i retract my downvote? excuse me sir ..... – user153330 Feb 4 at 22:43

I don't see why you should need variables for things known to be true or false.

Let $P,Q,R,S$ be as you chose them to be. You are given that $P$ and $S$ are true (final statements). So your first two statements become "If true and $Q$ then $R$" respectively "If true then not $R$". With $1,2$ denoting the simplified first two statements you get:

Q R 1 2

T T T F
T F F T
F T T F
F F T T


You need statements $1$ and $2$ to be true, so the last line must apply: unemployment does not go up and there will not be a recession.

You could alternatively pursue a bit further and remark that you second statement "If true then not $R$" just means "not $R$" (or "$R$ is false"), so that you have another known value, and can restrict to the second and fourth lines of the table while dropping the columns $R$ and $2$, giving

Q 1
T F
F T


This shows $Q$ is false as claimed. You could simplify statement $1$ to "not $Q$", but that would not leave much of a truth table, while you were asked to use them.

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