Propositional logic
Rules of Replacement
Material implication
p ⊃ q :: ∼p v q
2.How do you get From
p ⊃ q
to
∼p v q
?
My book doesn't show the process/prove.
In medieval logic, a disinction was made between material and formal consequence.
" John is a pianist, therefore John is a musician" : a material consequence, for the validity of the consequence depends on the semantic content ( matter) of the terms involved; it is not the case that, formally, " J is a P" implies " J is an M".
" All pianists are musicians, therefore, those who are not musicians are not pianists" : a formal consequence; the reasoning still holds when you replace the particular terms " pianist"/ " " musician" by variables.
A " consequence " ( in the medieval sense) is a reasoning ( with a " therefore" in it). This is not the same thing as an implication , whch only has an " if ... then " in it ( psychologically, an implication is a judgment, an assertion, not an inderence).
However there is an analogy between " material consequence" and " material implication".
A material implication only holds ( when it holds ) in virtue of the factual truth values of the propositions involved in it ( in the same way a material consequence holds in virtue of the semantic value of its terms).
Note : he truth value of a proposition is its semantic value understood as its denotation.
" If Biden is President of the USA then Harris is Vice-President".
This is a true material implication simply because the antecedent is true and the consequent is also true.
Saying that ths implicaion is true does not mean that Harris has to be Vice-President in case Biden is President. It simply means that, as a matter of fact, it is the case that both sentences are true, meaning that it is not the case that the first is true while the second is false.
Logical implication is different from material implication. The proposition " if P the Q " logically implies " if not-P, then not-Q", whatever the truth value of P and Q may be.
As to the question " why is P--> Q equivalent to ~P v Q?" . One could simply say that mplication is a derived connective, a connective that is defined in this way using primary connectives , namely negation and disjunction. Under this respect, the why question has no answer. But, under another point of view, one can say that implication is independently defined by its truth table, and that this truth table explains the equivalence.
Let me try to improve my answer to the second part of your question.
(1) One way to explain why P--> Q is equivalent to ~P v Q is simply to say : " I define a new connective, which I will call implication , P --> Q , as an abbreviation of ~P v Q." In that case, the two expressions are equivalent * by definition* and there is no aswer as to the "why" question : you cannot be wrong when you define the meaning of a term.
Note : in the same way one can say : I define A-B ( substraction ) as an abbreviation for " A + negative B". One cannot possibly be wrong when defining substraction, it's simply a definition. ( This is actually how substraction is defined in mathematics.)
(2) A second way is to say : I have 16 possible binary connectives, one of them has the truth table TT , TF, FT, FF. This connective I will call " material implication" . Once this is done, I notice that it has exactly the same truth table as the expression ~P v Q , and I call P --> Q and ~P v Q " equivalent". This is the semantic way ( since , through the truth tables, you make use of the possible meanings - namely the possible denotations/ truth values - of the sentences.)
(3) A third way is to say. I define P--> Q as ~ (P & ~Q). Then , using rules of transformation , I prove that P --> Q is equivalent to ~P v Q. This is the syntactic way ( since you only and blindly manipulate symbols according to rules, without referring to their possible denotations / truth values).
Derivation
P--> Q
is equivalent , by definition, to : ~ (P & ~Q)
which in turn is equivalent ( by De Morgan's) : ~P v ~ ~ Q
whic in turn is equivament ( by double negation) to : ~ P v Q.
Bertrand Russell coined the term "material implication", I assume he was thinking of "material" not in the sense of "relating to matter" but rather in a sense more like materiality in law, which basically just means something is considered relevant to the case at hand, as in the phrases material witness and material evidence. The answer by @FloridusFloridi covers the historical background of this notion of the "material" consequence of some statement of fact, as opposed to a "formal" consequence (where a formal consequence is one that doesn't depend on the specific meaning of any of the terms or in empirical facts about the entities named by those terms, only on the logical form of the statement and the statement expressing the consequence).
As for why it's equivalent to ∼p v q, that's just because it's defined to have the same truth table. It's not as if Russell was taking some prior natural-language notion of "implication" and proving that it should have this truth table--he did have some conceptual arguments for thinking material implication together with formal implication captured most of the ordinary-language notion of implication, see his paper "The Theory of Implication" and the paper 'Russell's Notion of Implication', but this was not any sort of logical proof. (Also note that on p. 161 of 'The Theory of Implication' Russell acknowledges that material implication doesn't capture all natural-language notions of implication, writing 'The meaning to be given to implication in what follows may at first sight appear somewhat artificial; but although there are other legitimate meanings, the one here adopted is, if I am not mistaken, very much more convenient than any of its rivals.') And later philosophers have pointed out that the indicative conditional of natural language can have other well-defined meanings not covered by either formal implication or material implication, like modal implications ('if A, then B' can be interpreted to mean 'in any possible world where A is true, B is also true' for some set of possible worlds, see for example the discussion of Lewis's analysis in terms of the set of 'closest possible worlds' on p. 10 of this paper), so in retrospect I think it's better to think of "material implication" as just an arbitrary name for one of the 16 possible binary logical operators.
A binary logical operator with this truth table is an especially useful one because it gives a natural way to translate logical statements of the form "all A are B" (which are commonly used in classical Aristotelian syllogisms) into first-order logic using material implication along with the universal quantifier, see my answer here for more on this point.
If you want to see a proof using only the simplest set of inference rules for first-order logic, see this chapter from the book A First Course in Logic, it gives 8 basic inference rules on p. 296, then on p. 300-301 they show how these can be used to derive De Morgan's laws, then on p. 302 they use the 8 basic rules along with De Morgan's laws to show that ∼p v q can be derived from p -> q. As for doing the reverse and deriving p -> q from ∼p v q, that's fairly simple if you allow the rule "double negation introduction" (this rule is derived from the 8 basic rules on p. 299 of the book chapter above):
1|_ ∼p v q Premise
2| |_ p Assumption
3| | ~~p 2 Double Negation Introduction
4| | q 1,3 Disjunctive Syllogism
5| p -> q 2-4 Conditional Proof
2.How do you get From p ⊃ q to ∼p v q
i was hoping for a proof in Natural Deduction style
It's a proof by contradiction.
1|_ p ⊃ q Premise
2| |_ ~(∼p v q) Assumption
3| | |_ p Assumption
4| | | q 1,3 Conditional Elimination
5| | | ∼p v q 4 Disjunction Introduction
6| | | # 5,2 Negation Elimination (using # as contradiction)
7| | ~p 3-6 Negation Introduction
8| | ∼p v q 7 Disjunction Introduction
9| | # 8,2 Negation Elimination
10| ~~(∼p v q) 2-9 Negation Introduction
11| ∼p v q 10 Double Negation Elimination
The converse is a proof by cases
1|_ ∼p v q Premise
2| |_ p Assumption
3| | |_ ~p Assumption (left case)
4| | | # 2,3 Negation Elimination
5| | | q 4 Explosion
| | +
6| | |_ q Assumption (right case)
7| | q 1,3-5,6-6 Disjunction Elimination
8| p ⊃ q 2-7 Conditional Introduction
