# How to prove logical consequence?

How do you guys prove for logical consequence? My teacher said a truth table can be done. But I don't understand how to infer from truth tables to establish logical consequences.

The definition in my course book says:

Let F1 and F2 be two formulae. F2 is a logical consequence of F1, we write F1 ⊧ F2 iff all models of F1 are models of F2.

I don't understand what it is being by "all models of F1 are models of F2", i keep thinking about logical equivalence for this statement, I can't discern between the two - logical equivalence and logical consequence. Hence, I have no idea on how to use the truth table to answer questions regarding logical consequence.

Can you guys help to shed light on this issue? Thank you.

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## 2 Answers

In the context of propositional logic, a model is simply an assignment of truth values to the basic propositional variables. The statement $p\land q$ is true in any model that satisfies both $p$ and $q$, i.e., in which $p$ and $q$ are both true; the statement $p\lor q$, on the other hand, is true in any model that satisfies at least one of $p$ and $q$. This information is precisely what you’re collecting when you write out a truth table.

To see whether a formula $F_2$ is a logical consequence of a formula $F_1$, write out their joint truth table. Find the rows in which $F_1$ is true; those are essentially the models that satisfy $F_1$. Check to see whether $F_2$ is also true in all of those rows/models; if it is, then every model that satisfies $F_1$ automatically satisfies $F_2$, and $F_2$ is therefore a logical consequence of $F_1$. Here’s an example.

\begin{align*} &F_1:\quad\big(p\to(q\lor r)\big)\land\lnot r\\ &F_2:\quad p\to q \end{align*}

Here’s the combined truth table:

$$\begin{array}{c|c} p&q&r&q\lor r&p\to(q\lor r)&\lnot r&p\to(q\lor r)\big)\land\lnot r&p\to q\\ \hline T&T&T&T&T&F&F&\color{blue}{T}\\ \color{green}{T}&\color{green}{T}&\color{green}{F}&T&T&T&\color{red}{T}&\color{blue}{T}\\ T&F&T&T&T&F&F&F\\ T&F&F&F&F&T&F&F\\ F&T&T&T&T&F&F&\color{blue}{T}\\ \color{green}{F}&\color{green}{T}&\color{green}{F}&T&T&T&\color{red}{T}&\color{blue}{T}\\ F&F&T&T&T&F&F&\color{blue}{T}\\ \color{green}{F}&\color{green}{F}&\color{green}{F}&F&T&T&\color{red}{T}&\color{blue}{T} \end{array}$$

The $T$’s for $F_1$ are in red, and those for $F_2$ are in blue. As you can see, in every row in which $F_1$ is true, $F_2$ is also true; the truth values of $p,q$, and $r$ in those rows are in green. $F_1$ holds in precisely those models in which $p$ and $q$ are true and $r$ is false, or $q$ is true and $p$ and $r$ are false, or all three are false. And in all such models $F_2$ holds as well, so $F_2$ is a logical consequence of $F_1$.

Note, though, that $F_2$ holds in some models in which $F_1$ does not hold, e.g., those in which $p,q$, and $r$ are all true. Thus, $F_1$ is not a logical consequence of $F_2$, and therefore $F_1$ and $F_2$ are not logically equivalent.

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Great answer as always Brian Scott, I always know to look for answers by you! I just started my first graduate logic course and was intimidated by the abstract notation at first (S-interpretations and all that jazz) but after reading through this I see that the abstraction for a similar question is just as easy as the more specific undergrad version of the question! Thank you for always formulating such comprehensive answers. – Clark Zinzow Sep 30 '13 at 19:05
@Clark: You’re welcome, and thank you. – Brian M. Scott Sep 30 '13 at 19:10
Scott, Thanks for your through explanation . This really resolved my doubts regarding "Logical Equivalence" and "Logical Consequence". Thank you. – user98322 Oct 2 '13 at 22:53
@Vijay: You’re welcome; I’m glad that it was so helpful. – Brian M. Scott Oct 3 '13 at 5:29

Logical equivalence goes in both directions; consequence goes one way only.

For example, in the elementary theory of groups, we can let $F_1$ be $\forall x:x*x=e$ and $F_2$ be $\forall x:x*x*x*x=e$.

Then $F_2$ is a consequence of $F_1$, because every group that satisfies $F_1$ also satisfies $F_2$. But they are not equivalent because there are groups where $F_2$ is true but $F_1$ isn't -- such as the cyclic group of order 4.

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