Understanding an argument from How to Prove It

Question

I am going through the book "How to prove it". But I am not quite able to wrap my head around this question from Velleman, Daniel J.. How to Prove It (A Structured Approach) (S.56). Cambridge University Press, which I am supposed to analyze with the help of a truth table.

Either sales or expenses will go up. If sales go up, then the boss will be happy. If expenses go up, then the boss will be unhappy. Therefore, sales and expenses will not both go up.

The answer to the question is that the reasoning is correct. I can sort of see how this is supposed to be true. Because the Boss cannot be both Happy and Unhappy at the same time only one of the two things can occur.

I use the following letters to denote the different statements:

• Sales go up (S)
• Expenses go up (E)
• Boss is Happy (H)
• Boss is Unhappy (¬H)
• S→H (If S then H)
• E→¬H (If E then not H)

S	E	H	¬H	S→H	E→¬H
F	F	F	T	T	F
F	T	F	T	T	F
T	F	F	T	T	F
T	T	F	T	T	T
F	F	T	F	T	F
F	T	T	F	F	F
T	F	T	F	T	F
T	T	T	F	F	T

I think there is something subtle I don't understand about conditionals. I hope someone can help me.

Answer 1

Either sales or expenses will go up. If sales go up, then the boss will be happy. If expenses go up, then the boss will be unhappy. Therefore, sales and expenses will not both go up.

• Sales go up (S)
• Expenses go up (E)
• Boss is Happy (H)

A valid argument is one whose conclusion is a logical consequence of its premises; so, in propositional logic, a valid argument is one whose corresponding conditional is a tautology. Thus, to mechanistically check whether the given argument is valid, we can construct the truth table of $$\Big((S\lor E) \quad\land\quad (S→H) \quad\land\quad (E→¬H)\Big)\quad\to\quad\lnot(S\land E)$$ then inspect its main column; every F in this column is a scenario in which all-true premises coexist with a false conclusion; this column being entirely Ts means that this argument's premises being true forces its conclusion to hold.

\begin{array}{ccc|c@{}c@{}c@{}c@{}ccc@{}ccc@{}ccc@{}c@{}ccc@{}ccc@{}cc@{}c@{}c@{}ccc@{}cc@{}ccc@{}c@{}c@{}c} E&H&S&(&((S&\lor&E)&&\land&&(S&\rightarrow&H))&&\land&&(E&\rightarrow&\lnot&H)&)&\rightarrow&&\lnot&(S&\land&E)&&\\\hline 1&1&1&&1&1&1&&1&&1&1&1&&0&&1&0&0&1&&\mathbf{1}&&0&1&1&1&&&\\ 1&1&0&&0&1&1&&1&&0&1&1&&0&&1&0&0&1&&\mathbf{1}&&1&0&0&1&&&\\ 1&0&1&&1&1&1&&0&&1&0&0&&0&&1&1&1&0&&\mathbf{1}&&0&1&1&1&&&\\ 1&0&0&&0&1&1&&1&&0&1&0&&1&&1&1&1&0&&\mathbf{1}&&1&0&0&1&&&\\ 0&1&1&&1&1&0&&1&&1&1&1&&1&&0&1&0&1&&\mathbf{1}&&1&1&0&0&&&\\ 0&1&0&&0&0&0&&0&&0&1&1&&0&&0&1&0&1&&\mathbf{1}&&1&0&0&0&&&\\ 0&0&1&&1&1&0&&0&&1&0&0&&0&&0&1&1&0&&\mathbf{1}&&1&1&0&0&&&\\ 0&0&0&&0&0&0&&0&&0&1&0&&0&&0&1&1&0&&\mathbf{1}&&1&0&0&0&&& \end{array}

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Addendum

Thanks. That's a really detailed answer. So, because of the Therefore in the last sentence, we have to prove that the main conditional, rather than the conclusion $¬(S \land E),$ is a tautology?

Yes, exactly. A valid argument does not require its conclusion to be valid (i.e., true regardless of whether its premises are all true) or even true in your particular scenario—just that its conclusion be true in every scenario in which its premises are all true.

For example, the argument "I am both tall and short; therefore, I am both tall and short" is valid yet its conclusion is a contradiction.

In your example argument, we don't care that its conclusion $¬(S \land E)$ is false in scenarios 1 and 3, since in these rows its premises are not all true.