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1.4 Arguments
Definition
A valid argument is a finite set of propositions P 1 , ..., Pr called premises, together with a proposition C, the conclusion, such that the propositional form (P 1 ∧ P 2 ∧ ... ∧ Pr) → C is a tautology.
We say C follows logically from, or is a logical consequence of the premises. We write P 1 , ..., Pr C. The symbol is called the turnstile.
Definition
If an argument in not valid we say that it is invalid.
First Method to prove validity
Example 19
Let Q 1 = “John graduates”
Q 2 = “Mary graduates” Q 3 = “John gets a job” Q 4 = “Mary gets a job” Q 5 = “Mary earns money”
(i) Consider the following argument:
“If John graduates then he gets a job”. “John graduates”. “Therefore John gets a job”.
To see the “form” of this argument we symbolize it as Q 1 → Q 3 , Q 1 ` Q 3.
Now we check that ((Q 1 → Q 3 ) ∧ Q 1 ) → Q 3 is a tautology:
Q 1 Q 3 Q 1 → Q 3 (Q 1 → Q 3 ) ∧ Q 1 ((Q 1 → Q 3 ) ∧ Q 1 ) → Q 3
T T T T T
T F F F T
F T T F T
F F T F T
This tautology shows that the argument is valid.
Note Another “instance” of this argument follows if we set Q 1 = “2 < 1” and Q 3 = “3 < 2”. The argument then reads:
“If 2 < 1 then 3 < 2” “2 < 1” “Therefore, 3 < 2”.
This is still valid though some of the propositions, i.e. 2 < 1 for instance, are false.
Example 19 continued. (ii) Consider the following argument:
“If Mary graduates then she gets a job”. “Mary does not get a job”. “Therefore Mary does not graduate”.
Symbolized, this becomes Q 2 → Q 4 , (¬ Q 4 ) ` (¬ Q 2 ).
*Now check that ((Q 2 → Q 4 ) ∧ (¬ Q 4 )) → (¬ Q 2 ) is a tautology.
Q 2 Q 4 Q 2 → Q 4 ¬ Q 4 (Q 2 → Q 4 ) ∧ (¬ Q 4 )(≡ A) ¬ Q 2 A → (¬ Q 2 )
T T T F F F T
T F F T F F T
F T T F F T T
F F T T T T T
(iii) Consider the following argument:
“Either Mary or John graduate”. “John does not graduate”. “Therefore Mary graduates”.
Symbolized, this becomes Q 2 ∨ Q 1 , (¬ Q 1 ) ` Q 2.
*Now check that ((Q 2 ∨ Q 1 ) ∧ (¬ Q 1 )) → Q 2 is a tautology.
Q 1 Q 2 Q 2 ∨ Q 1 ¬ Q 1 (Q 2 ∨ Q 1 ) ∧ (¬ Q 1 )(≡ B) B → Q 2
T T T F F T
T F T F F T
F T T T T T
F F F T F T
false if C is false and P 1 ∧ P 2 ∧ ... ∧ Pr is true, i.e. all P 1 , P 2 ,...,Pr are true. So we never want to see a row in the truth table where all the premises are true and the conclusion false.
This observation gives a second way of checking that an argument is valid or not.
Example 20
Is p → q, q ∨ p ` (¬ q) ∨ (¬ p) valid?
We look at the truth table:
p q p → q q ∨ p (¬ q) ∨ (¬ p) T T T T F ← T F F T T F T T T T F F T F T
In the first line the conclusion is false, but all premises are true. Hence the argument is invalid.
This method requires fewer columns than in the first method. Is there an “instance” of this argument which is “obviously” invalid? Try looking in the “World of Mathematics”, for instance, choosing p ≡ “3 > 2” and q ≡ “2 > 1”. Then the argument becomes:
If 3 > 2 then 2 > 1, Either 3 > 2 or 2 > 1, Therefore, either 3 ≤ 2 or 2 ≤ 1. Both premises are true but the conclusion is false. On the basis that we never want a false conclusion to follow from true premises, this argument is invalid. What we have here is an example where p is True and q is True, which is the first line in the table, where we had seen the problem.
But be careful! Consider another instance. So let p ≡ “Stockport is a city” and q ≡ “Manchester is a city”. Then the argument becomes:
If Stockport is a city then Manchester is a city, Either Manchester or Stockport is a city, Therefore, either Manchester is not a city or Stockport is not a city. If I tell you that Manchester is a city but Stockport is not a city then you can check that all the propositions in this argument are true. But the argument is still invalid. It is a case of the conclusion, though true, not following logically from the true premises.
Example 21 Is p → (s → (¬ r)), p → r, p ` ¬ s valid?
p r s ¬ r (s → (¬ r) p → (s → (¬ r)) p → r p ¬ s T T T F F F T T F T T F F T T T T T T F T T T T F T F T F F T T T F T T F T T F F T T F F F T F F T T T F T F F T T T T T F F F F F T T T T F T
We look at each row in turn. We look to see if on any row we have a case of all the premises being true with the conclusion false. For instance in the first row we see that premises are F, T, T and the conclusion F. This is allowable. By checking each row we see that each row is allowable, that is, we never have a case of all premises true with the conclusion false. Hence the argument is valid. This method is straightforward. In fact, a machine can do it.
Note I have given here two methods for using a truth table to check whether P 1 , ..., Pr ` C is valid or not.
Do not mix up these methods! In the first method use a truth table to work out the truth values of (P 1 ∧ P 2 ∧ ... ∧ Pr) → C, and hope that it is always true, i.e. a tautology.
In the second method construct a table containing a column for each of the P 1 , P 2 , ... up to Pr along with C and hope that there is no row with all the Pi true and C false.
If an argument is invalid there is sometimes a quick method of showing this.
Example Show that
(p ∨ q) → s, q → s ` s
is invalid. We do this by trying to make the conclusion false and the premises all true.
