London Ed refers us to Understanding Arguments: an Introduction to Informal Logic, Robert Fogelin and Walter Sinnott-Armstrong, and provides this quotation:
Perhaps a bit more surprisingly, our definitions allow 'roses are red and roses are red' to be a substitution instance of 'p & q'. This example makes sense if you compare it to variables in mathematics. Using only positive integers, how many solutions are there to the equation 'x + y = 4'? There are three: 3+1, 1+3, and 2+2. The fact that '2+2' is a solution to 'x + y = 4' shows that '2' can be substituted for both 'x' and 'y' in the same solution. That's just like allowing 'roses are red' to be substituted for both 'p' and 'q', so that 'roses are red and roses are red' is a substitution instance of 'p & q' in propositional logic.
In general, then, we get a substitution instance of a propositional form by uniformly replacing the same variable with the same proposition throughout, but different variables do not have to be replaced with different propositions. The rule is this:
Different variables may be replaced with the same proposition [Ed: Let's call this the London rule], but different propositions may not be replaced with the same variable.
Suppose I am given the task of determining whether the conditional English sentence 'If roses are red, then roses are red' is a tautology, a contradiction, or a contingency. How do I proceed?
Step One is translation, or encoding. Let upper case letters serve as placeholders for propositions. Let '–>' denote the truth-functional connective known in the trade as the material or Philonian conditional. I write 'P –> P.'
Step Two is evaluation. Suppose for reductio that the truth value of 'P –>P' is false. Then, by the definition of the Philonian conditional, we know that the antecedent must be true, and the consequent false. But antecedent and consequent are the same proposition. Therefore, the same proposition is both true and false. This is a contradiction. Therefore, the assumption that conditional is false is itself false. Therefore the conditional is a tautology.
Now that obviously is the right answer since you don't need logic to know that 'If roses are red, then roses are red' is a tautology. (Assuming you know the definition of 'tautology.') But if if Fogelin & Co. are right, and the 'P –>Q' encoding is permitted, then we get the wrong answer, namely, that the English conditional is a contingency.
I am assuming that if 'P–>Q' is a logical form of 'If roses are red, then roses are red,' then 'P –>Q' is a legitimate translation of 'If roses are red, then roses are red.' As Heraclitus said, the way up and the way down are the same. The assumption seems correct.
If I am right, then there must be something wrong with the mathematical analogy. Now there is no doubt that Fogelin and his side kick are right when it comes to mathematics. And I allow that what they say is true about variables in general. Suppose I want to translate into first-order predicate logic with identity the sentence, 'There is exactly one wise man.' I would write, '[(Ex)Wx & (y)(Wy –> x = y)].' Suppose Siddartha is the unique wise man. Then Siddartha is both the value of 'x' and the value of 'y.'
So different variables can have the same value. And they can have the same substituend. In the example, Siddartha is the value and 'Siddartha' is the substituend. But is a placeholder the same as a variable? I don't think so. Here is a little argument:
No variable is a constant
Every placeholder is an arbitrary constant
Every arbitrary constant is a constant
——-
No placeholder is a variable.
A placeholder is neither an abbreviation, nor a variable. It is an arbitrary constant. Thus the logical form of 'Al is fat' is Fa, not Fx. Fa is a proposition, not a propositional function. 'F' is a predicate constant. 'a' is an individual constant. We cannot symbolize 'Al is fat' as Fx. For Fx is not a proposition but a propositional function. If 'a' were not an arbitrary constant, then Fa would not depict the logical form of 'Al is fat,' a form it shares with other atomic sentences.
Here is another argument:
Every variable is either free or bound by a quantifier
No placeholder is either free or bound by a quantifier
——-
No placeholder is a variable.
Here is a third argument:
Every variable has a domain over which it ranges
No placeholder has a domain over which it ranges
——-
No placeholder is a variable.
A fourth argument:
There is no quantification over propositions in the propositional calculus
——-
There are no propositional variables in the propositional calculus
If there are no propositional variables in the propositional calculus, then the placeholders in the propositional calculus cannot be variables
——-
The placeholders in the proposition calculus cannot be variables.
Punchline: because placeholders are not variables, the fact that the different variables can have the same value and the same substituend does not show that different placeholders can have the same substituend. 'If roses are red, then roses are red' does not have the logical form 'P –>Q' and the latter form does not have as a substitutution-instance 'If roses are red, then roses are red.'
As I have said many times already, one cannot abstract away from the fact that the same proposition is both antecedent and consequent.
What one could say, perhaps, is that 'P –> P' has the higher order form 'P –> Q.' But this latter form is not a form of the English sentence but a form of the form of the English sentence.
Ed can appeal to authority all he wants, but that is an unphilosophical move, indeed an informal fallacy. He needs to show where I am going wrong.
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