Compound sentences of the form "If P, then Q" can perhaps be best understood as a promise. Truth for the whole compound sentence is decided on the basis of whether you think the promise is kept. Here P and Q are themselves sentences, each of which is true or false (but neither is both true and false). Think of the following promise:

If there is an elephant in the library, then I'll give you 100 dollars.Suppose there really is an elephant in the library and I give you 100 dollars. Then I've kept my promise and the whole "if...then" sentence is true.

Suppose there really is an elephant in the library, but I don't give you any money. Then I've broken my promise and the whole "if...then" sentence is false.

Now suppose that there is no elephant in the library and I don't give you any money. I've kept my promise and the whole "if...then" is true.

What happens if there is no elephant in the library, but I give you 100 dollars anyway? When I ask students this case in class, it often splits 50-50. Half think I've broken the promise. Half think I've kept it. ORDINARY ENGLISH IS NOT CLEAR ABOUT THIS CASE. In the technical sciences (mathematics, engineering, etc.) we had to make a choice to read it one way or the other (remember, computers are complete idiots and you must be utterly exact with them). IN THE TECHNICAL SCIENCES, THE PROMISE IS VIEWED AS BEING KEPT. "IF P, then Q" is true when P is false and Q is true. Think of it this way: you're allowed to be generous in a promise and honest at the same time.

The truth table below formalizes the discussion above. T stands for true and F stands for false. Note the third line, which is the case where ordinary English is not clear.

P | Q | If P, then Q |
---|---|---|

T | T | T |

T | F | F |

F | T | T |

F | F | T |

P | Q | If P, then Q | If Q, then P |
---|---|---|---|

T | T | T | T |

T | F | F | T |

F | T | T | F |

F | F | T | T |

What happens if you interchange (commute) P and Q in an "if...then"? Generally, you get a sentence with a different truth. For example, here is a true "if...then": if an animal is a bird, then it is a warm-blooded animal (this sentence is always true). However, the commuted "if...then" is false (meaning, sometimes false; not reliably true): if an animal is warm-blooded, then it is a bird. This is false about people for example: people are warm-blooded but they are not birds. A side-by-side comparison of truth tables for "if P, then Q" and "if Q, then P" show that they do not mean the same thing in the second and third rows:

P | Q | If P, then Q | not(P) | (not(P)) or Q |
---|---|---|---|---|

T | T | T | F | T |

T | F | F | F | F |

F | T | T | T | T |

F | F | T | T | T |

This last truth table proves that

The vast majority of theorems have the form "if P, then Q". Here are three common styles of proof. First, the writer assumes that P is true. Somehow, using P and other theorems, the writer derives Q and stops. What the writer is skipping is the easy case: suppose P is false. Well, then Q doesn't matter. According to the truth tables above, the whole "if...then" is true when the "if" part is false. So the writer assumes that you know this easy part and doesn't bother to mention it. The writer has eliminated the one case where "if P, then Q" is false: P is true and Q is false.

A second style of proof is begins by assuming that "if P, then Q" is false and derives a contradiction from that. In the truth tables above, there is only one case where "if P, then Q" is false: namely, P is true and Q is false. Typically, the writer will skip to this combination (assume P is false and Q is true) and derive his contradiction from those two statements and then stops. The writer assumes that you know when "if P, then Q" is false. The writer also assumes that you know getting a contradiction from assuming "if P, then Q" is false means that "if P then Q" must be therefore be true. Why waste words on explaining the obvious? After awhile, it does indeed become obvious.

A third style proceeds by a chain of "if...then"'s. The writer explains that "if P, then S". Then the writer explains that "If S, then Q". Now the writer stops. It's supposed to be obvious that "if P, then Q" is now proven (and to most people it is). The chain can be quite long, but principle is the same: if P then S1; if S1 then S2; if S2 then S3;....; if S99, then Q. Therefore, if P, then Q. Let us review a short chain carefully: if P then S; and if S, then Q. How does this prove "if P, then Q"? Suppose that P is true. Look at the truth table for "if P then S"; for this "if...then" to be true with P being true, S has to be true. Now play the same trick with "if S then Q": for this "if...then" to be true with S being true, Q has to be true. So, when P is indeed true, so is Q. The combination of P is true with Q is false DOES NOT OCCUR. Since this is the only time "if P then Q" is false, we know that "if P then Q" is true.

The sentence "If [(if P, then Q) and (if Q, then R)], then (if P, then
R)" captures the principle of the previous paragraph. It is an example of
a **tautology**, a sentence which is always true regardless
of the truth of P, Q, and R. Here is a table that establishes this
tautology:

P | Q | R | if P, then Q | if Q, then R | [(if P, then Q) and (if Q, then R)] | if P, then R | If[(if P, then Q) and (if Q, then R)], then (if P, then R) |
---|---|---|---|---|---|---|---|

T | T | T | T | T | T | T | T |

T | T | F | T | F | F | F | T |

T | F | T | F | T | F | T | T |

T | F | F | F | T | F | F | T |

F | T | T | T | T | T | T | T |

F | T | F | T | F | F | T | T |

F | F | T | T | T | T | T | T |

F | F | F | T | T | T | T | T |

By a principle of induction, one can extend this transitive principle to any number of sentences.

Another, and rather horrendous approach to the previous principle, is to translate each "if...then" occurrence in it to the equivalent "not... or" and then simplify rather deviously:

if[(if P, then Q) and (if Q, then R)], then (if P, then R) |

means the same as |

not[(if P, then Q) and (if Q, then R)] or (if P, then R) |

which means the same as (see "not" applied to an "and" sentence) |

{[not(if P, then Q)] or [not(if Q, then R)]} or (if P, then R) |

which means the same as (see "not" applied to an "if...then" sentence ) |

{[P and (not(Q))] or [Q and (not(R))]} or (if P, then R) |

which means the same as (see "or" distributes over "and") |

{[[P and (not(Q))]or Q] and [[P and (not(Q))] or (not(R)) ]} or (if P, then R) |

which means the same as (because "or" is commutative, see "or") |

{[Q or [P and (not(Q))]] and [(not(R)) or [P and (not(Q))]]} or (if P, then R) |

which means the same thing as (because "or" distributes over "and") |

{[(Q or P) and (Q or (not(Q)))] and [(not(R)) or [P and (not(Q))]] } or (if P, then R) |

which, because Q or (not(Q)) is always true, means the same thing as |

{(Q or P) and [(not(R)) or [P and (not(Q))]] } or (if P, then R) |

which, because "if...then" means the same as a "not .... or", means the same thing as |

{(Q or P) and [(not(R)) or [P and (not(Q))]] } or [(not(P)) or R] |

which, because "or" is commutative, means the same thing as |

[(not(P) or R] or {(Q or P) and [(not(R)) or [P and (not(Q))]] } |

which, because "or" distributes over "and", means the same thing as |

{ [(not(P)) or R] or (Q or P)} and {[(not(P)) or R] or [(not(R)) or [P and (not(Q))]] } |

which, because [((not(P)) or R) or (Q or P)] is always true ("not(P) or P" is part of this larger "or" and always true, means the same as |

{[(not(P)) or R] or [(not(R)) or [P and (not(Q))]]} |

which is always true because it means the same as "not(R) or R" "or"-ed with other sentences and "not(R) or R" is always true. |

Go to an overview of logic. |