Why is the **material implication of classical logic** (also known as material conditional or material consequence), p -> q, defined to be false only when its antecedent (p) is true and the consequent (q) is false? Here is an informal way to think about it.

You could view logic as metamathematics, a language designed to talk about mathematics. Logic as the “hygiene”, the grammar and syntax of mathematics.

In the language of classical logic every proposition is either true or not true, and no proposition can be both true and not true. Now what if we want to express the natural language construction “If…then…” in this language? Well, there are exactly sixteen possible truth functions of two inputs p and q (since there are 2^2 inputs and (2^2)^2 ways to map them to outputs). And the candidate that best captures the connotations of what we mean by “If…then…” is the definition of material implication. Here is why.

By stating that p -> q is true we want to indicate that the truth of q can be inferred from the truth p, but that nothing in particular can be inferred from the falsity of p. And this is exactly the meaning captured by the material conditional:

p | q | p->q |
---|---|---|

T | T | T |

T | F | F |

F | T | T |

F | F | T |

First, when “If p, q” is true, and we also know that p is true, then we want to be able to infer q. In other words, if we claim that if p is true then q is true, then if p is indeed true, q should be true as well. This basic rule of inference has a name, it is called modus ponens.

Second, if we claim “If p, q”, then if p is false, we did not say anything in particular about q. If p is false, q can either be true or false, our claim “If p, q” is still true.

But notice that it is not possible to capture all notions of what we colloquially mean by “If…then…” statements as a two-valued truth function.

It is for example possible to make meaningless statements such as “If grass is red then the moon if made of cheese.” This is however unproblematic under the assumption that logic is an idealized language, which is adequate for mathematical reasoning. Since we are mainly interested in simplicity and clarity. Under this assumption, such nonsense implications are analogous to grammatically correct but meaningless sentences that can be formed in natural languages, such as “Colorless green ideas sleep furiously“.

To demonstrate its adequacy for mathematics, here is a mathematical example:

If n > 2 then n^2 > 4.

We claim that if n is greater than 2 then its square must be greater than 4. For n = 3, this is obviously true, as we claimed. But what about n smaller than 2? We didn’t say anything in particular about n smaller than 2. Its square could be larger than 4 or not. And indeed, n = 1 and n = -3 yield a false, respectively true, consequent. Yet the implication is true in both cases.

Intuitively more problematic are statements such as (p and not(p)) -> q, p and its negation imply q. Think about it this way. The previous implication is a tautology, it is always true. And you believe true statements. This however does not mean that you must believe that arbitrary q is true too (as long as you stay consistent), since in case of the falsity of the antecedent you are not making any particular claim about the truth of the consequent (q). And since the statement that p is true and false, p AND not(p), is always false — remember the principle of exclusive disjunction for contradictories, (P ∨ ¬P) ∧ ¬(P ∧ ¬P), requires that every proposition is either true or not true, and that no proposition can be both true and not true — q can be false without invalidating the implication.

Another way to look at p -> q is by interpreting it as “p is a subset of q”. Then if it is true that x *is an element of* p, then it must be true that it is also an element of q (since q contains p). However, if x is not an element p, then it might still turn out to be an element of q, since q can be larger than p.

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Here is a term I just learnt: **Extraneous solutions**.

Take for example the equation

A = B.

If you were to square both sides you would get

A^2 = B^2

or

A^2 – B^2 = 0.

Which is equal to

(A – B)(A + B) = 0 (by the difference of two squares).

Now the roots of this equation are the roots of the equations A = B and A = -B. This means that we generated an additional solution by squaring the original equation.

The reason for this is that squaring is *not* an injective fuction (injective means one-to-one, every element is mapped to one and only one unique element), it is not invertible. The function y = x^2 does not pass the horizontal line test. In other words, squaring preserves equality, if A = B then A^2 = B^2, but does not preserve inequality. It is not true that if A != B then A^2 != B^2, since both -1 and 1 are mapped to 1 when squared. Which means that both 1^2 = 1^2 and (-1)^2 = (1)^2 are solutions to the squared equations, while only one of them makes each pre-squared equation true.

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**Video:** http://youtu.be/Y53vDnNPiA4

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