A question about Transposition

Question

Propositional logic

Rules of Replacement

Transposition (Trans):

( p ⊃ q) :: (∼q ⊃ ∼p)

All i see here is some kind of a modus tollens case I don't see how did they get from

( p ⊃ q)

To

(∼q ⊃ ∼p)

or why it is considered an equivalent.

(I mean by a logical process or explanation not by truth table,even if it's true by the truth table, a truth table isn't a proof for something, there is no rational there only to notice something like equivalent and i don't think i can use a truth table as a proof, that is a very weak proof.)

Answer 1

For clarification, given the way you've written the question, I'm supposing that your question is about the rule of transposition in classical logic and that you are using ⊃ to represent material implication. The term 'transposition' has a different though related meaning in Aristotelian logic.

If you are struggling to understand why the rule is correct, you might like to start just by considering some examples. "If a thing is a bird then it has wings" has the same truth conditions as, "If a thing does not have wings, then it is not a bird". "If it rains on Friday the match is cancelled" has the same truth conditions as, "If the match was not cancelled, it didn't rain on Friday". "If the buyer did not make payment by the specified date, the contract was voided" has the same truth conditions as, "If the contract was not voided, the buyer made payment by the specified date".

Note that I'm speaking here of truth conditions, because conditionals in the real world can carry all kinds of additional meanings and implicatures. Material implication is a special kind of conditional that boils the conditional down to its bare truth conditions only. P ⊃ Q is equivalent to ¬(P ∧ ¬Q). It holds that the truth of P is a sufficient condition for the truth of Q, and the truth of Q is a necessary condition of the truth of P. Hence, if Q is false, by implication P is false.

It is not irrelevant or weak to demonstrate the equivalence by truth table, and it is easily done. We can also demonstrate the equivalence using natural deduction. The exact form will depend on what rules you use, but if we allow ourselves a little freedom, then the proof is very simple:

1. P ⊃ Q              Premise
2. ¬(P ∧ ¬Q)          From 1, by equivalence of implication
3. ¬P ∨ Q             From 2, by de Morgan 
4. ¬Q                 Assumption
5. ¬P                 From 3, 4, by disjunctive syllogism 
6. ¬Q ⊃ ¬P            From 4, 5, discharging the assumption in 4. 

The converse can be proved in the same way with the use of double negation elimination.

It is worth noting that not all logics have this rule. For example, in intuitionistic logic, P → Q entails ¬Q → ¬P, but the converse does not hold. Also, transposition, and the corresponding rule of contraposition, does not hold in probability logic, because if the conditional is uncertain, a high value for P(B|A) does not entail a high value for P(¬A|¬B). Also, contraposition does not hold in general for counterfactual conditionals.

Answer 2

A => B or "If A is true, B must be true", means that if B is found to be false, A must be false too because if it was true B would be too (and A can only be true or false according to the excluded middle principle).

You can picture it using the equivalence of implication with inclusion: figure 2 circles A and B, with A entirely inside of B. This represents "A => B": if a point is inside A it must be inside B. But you can also see that there is no point outside of B that is inside of A, therefore "~B => ~A".

By the way, truth tables are a perfectly valid demonstration.

Answer 3

To prove logical equivalence syntactically you need to show that $\phi \vdash \psi$ and $\psi \vdash \phi$, where $\psi$ and $\phi$ are wffs.

To do that, start by taking $\phi$ as a premise and then deriving $\psi$. Followed by taking $\psi$ as a premise and then deriving $\phi$.

Another way to do it is by showing $\vdash \phi \to \psi$ and $\vdash \psi \to \phi$, which allows us to infer $\vdash \phi \leftrightarrow \psi$.

Modus Tollens, conditional proof, and how to handle double negation is all that's required for the proof.