include $_SERVER["DOCUMENT_ROOT"].'/think/module.php'; section('Logical status'); ?> Armed with truth-tables we can now use them to classify WFFs in SL according to their different logical status :
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Question 2 Determine whether these WFFs are tautological, contingent, or inconsistent:
- Is it true that every WFF is either a tautology, inconsistent, or contingent? popupbox('Yes, as can be seen in the definition.'); ?>
- Is every tautology consistent? popupbox('Yes.'); ?>
- All consistent WFFs are contingent. popupbox('No.'); ?>
- If a WFF is consistent, then there is at least one assignment where it is false. popupbox('An inconsistent WFF is one that can never be true. A tautology that is always true is consistent.'); ?>
section('Consistency'); ?> Apart from talking about the properties of individual WFFs, we can also use truth-tables to identify logical relations between WFFs in SL.
- P popupbox('contingent'); ?>
- ((P&~P)∨Q) popupbox('contingent'); ?>
- (~R∨(P→((Q→R)&(~S∨R)))) popupbox('tautology'); ?>
- (~Q→Q) popupbox('contingent'); ?>
- ((~P&~Q)↔(P∨Q)) popupbox('inconsistent'); ?>
- ((P&~P)&Q) popupbox('inconsistent'); ?>
Earlier we talked about a single WFF being consistent or inconsistent. Actually we can also talk about consistency as the property of a set of one or more WFFs. A set of WFF is said to be consistent (with each other) when there is at least one assignment of truth-value under which all the WFFs in the set are true. Otherwise the set of WFFs is inconsistent - there is not even one single assignment of truth-values that would make all the WFFs true. For example, the set containing only "(P→Q)", "~P" and "(P∨Q)" is a consistent set of SL WFFs. This set is consistent since there is an assignment of truth value under which all 3 WFFs are true, namely, when P is false and Q is true. The set containing only "((P&Q)→~R)", "(~P∨~Q)", and "~R" is inconsistent. There is no assignment of truth values which makes all 3 WFFs true.
ex(); ?>
section('Entailment'); ?> Suppose we have a WFF φ and a set of (one or more) WFF ψ1...ψn. We define entailment as follows :
- If a set of WFFs is inconsistent, would the set become consistent by adding more WFFs to it? popupbox('No.'); ?>
- If you have a set containing only tautologies, is the set consistent? popupbox('Yes, of course.'); ?>
- If A is consistent with B, and B is consistent with C, does it follow that A must be consistent with C? popupbox('No, e.g. A="Bond is a spy", B="Bond likes Martini", C="Bond is not a spy".'); ?>
- If A is inconsistent with B, and B is consistent with C, does it follow that A must be inconsistent with C? popupbox('No, e.g. A="Bond is a spy", B="Bond is not a spy", C="Bond likes Martini".'); ?>
- If X, Y and Z form an inconsistent set of WFFs, then X and Y are inconsistent with each other. popupbox('No, e.g. X is "P", Y is "Q" and Z is "(R&~R)".'); ?>
- If X is an inconsistent WFF, and Y is an inconsistent WFF, then X is inconsistent with Y. popupbox('Yes.'); ?>
ψ1...ψn entail φ if and only if there is no assignment of truth-value under which ψ1...ψn are true and φ is false.So for example, ~P entails ~~~P. To show this we draw their truth-tables together :
Another example :
P | Q | (P↔Q) | ~P | ~Q |
T | T | T | F | F |
T | F | F | F | T |
F | T | F | T | F |
F | F | T | T | T |
You should be able to work out for yourself that these claims are true :
Entailment : ψ1...ψn entail(); ?> φ No Entailment : ψ1...ψn notentail(); ?> φEquivalently we might also say :
Notice that an argument in a formal system of logic is often called a sequent. So a valid argument in SL is called a "valid sequent", and an invalid argument in SL is called an "invalid sequent".
Now that we know what entailment is, we can easily prove these two theorems :
Theorem 1 : For any WFF φ, φ entail(); ?> φ.Proof : Since φ is identical to itself, there cannot be any assignment where φ is true and φ is false. So φ entails φ.
Theorem 2 : For any WFFs φ1 ... φn and any tautology ψ, φ1 ... φn entail(); ?> ψ.In other words, a tautology is entailed by any set of WFFs. Proof : Given that ψ is a tautology, it is true under all assignments. So there is no assignment where ψ is false and φ1 ... φn are true, whatever φ1 ... φn are. section('Logical equivalence'); ?> Logical equivalence can be defined in terms of entailment as follows :
For any WFFs φ and ψ, φ is logically equivalent to ψ if and only if φ entail(); ?> ψ and ψ entail(); ?> φ.We might use this symbol "≡" to express logical equivalence :
φ is logically equivalent to ψ : φ ≡ ψObviously, these claims are all true :
(P→Q) ≡ (~P∨Q)
(P&Q) ≡ ~(~P∨~Q)
(P∨Q) ≡ ~(~P&~Q)
(P↔Q) ≡ ((P→Q)&(Q→P))
section('Inconsistent WFFs entail everything'); ?> You would probably find it surprising that given our definition of entailment, inconsistent WFFs entail every WFF. So for example, the following sequent is valid :
P, ~P entail(); ?> QThis sequent is valid because there is no assignment under which P, ~P are true and Q is false. There is no such assignment simply because P, ~P cannot be true! By the same reasoning, the sequent will still be valid whatever WFF you replace Q with.
No doubt you might find it hard to accept that from an inconsistent sentence, everything follows. Here is a way of understanding why we might want to live with a definition of entailment that has such a consequence. First of all, presumably we want the following sequent to be valid :
The earth is round entail(); ?> Either the earth is round, or pigs can fly.On the other hand, we also want this sequent to be valid :
Either the earth is round, or pigs can fly. The earth is not round. entail(); ?> Pigs can fly.If these two sequents are valid, then surely we should also have a valid sequent when we put the two together :
The earth is round. The earth is not round. entail(); ?> Pigs can fly.Of course, you can replace "pigs can fly" with any statement you want. This shows that as long as we accept that (1) from any statement "P" we can conclude "(PvQ)", and (2) from "(PvQ)" and "~P" we can conclude "Q", then we would have to accept that "P" & "~P" entail all statements!
If you still find it strange that from an inconsistent sentence, everything follows, you might be interested in this (supposedly true) story about the famous philosopher / logician Bertrand Russell. He was asked the question, You mean from the statement 2+2=5 it follows that you are the Pope? Can you prove it? Russell said yes and then came up with this proof on the spot:ex(); ?> Are these statements true or false?Now, Russell continues, the Pope and I are two. Since two equals one, then the Pope and I are one. Hence I am the Pope.
- Suppose 2+2=5.
- Subtracting 2 from both sides we get 2=3.
- Transposing, we have 3=2.
- Subtracting 1 from both sides, we get 2=1.
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- If a contingent WFF X entails another WFF Y, then Y is also contingent. popupbox('No. Y can be a tautology.'); ?>
- If a contingent WFF X entails another WFF Y, then Y cannot be inconsistent. popupbox('True. Since X is contingent, it is possible for X to be true. When X is true, Y must also be true if X entails Y. So it is possible for Y to be true.'); ?>
- If a tautology X entails a WFF Y, then Y is also a tautology. popupbox('Yes. Since X is true under all assignments, and Y has to be true whenever X is.'); ?>
- If the conclusion of a sequent is inconsistent, then the sequent must be valid. popupbox('No! Not if the premises are consistent!'); ?>
- If A is inconsistent with B, and C entails B, does it follow that A must be inconsistent with C? Why? popupbox('Yes. If C is true, then B is also true, and so A cannot be true at the same time.'); ?>
- If an SL WFF X contains 2 sentence letters, then if a different SL WFF Y is logically equivalent to X, Y contains those 2 same sentence letters. popupbox('False. "((A&~A)&B)" is logically equivalent to "((C&~C)&D)".'); ?>
- If SL WFFs X and Y are logically equivalent, and the negation of Y is consistent, then the conjunction of X and Y is consistent. popupbox('False.'); ?>