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$$(A ∧ (A → B)) ⇒ B$$

I know that this is a tautology, but apart from setting up a truth-table I dont know how I would go about a formal proof in discrete mathematics. thanks

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Mauro ALLEGRANZA
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Iver Finne
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  • start from $A\to B=\lnot A\lor B$ then distribute $\land$ over $\lor$ and simplify. – zwim Feb 23 '21 at 12:07
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    It seems notations differ by location. What is the difference between $\rightarrow$ and $\Rightarrow$? – VIVID Feb 23 '21 at 12:09
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    It depends on what "environment" for formal proof you are using... – Mauro ALLEGRANZA Feb 23 '21 at 12:39
  • It's just one of the essential, self-evident and defining properties of logical implication. It shouldn't need a proof. Yes, I'm sure you can concoct a list of less self-evident axioms and derive this principle (Detachment or Modus Ponens) from those axioms. – Dan Christensen Feb 23 '21 at 16:18

3 Answers3

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$(A \land (A\rightarrow B)) \rightarrow B $ is true when $A \land (A\rightarrow B)$ is false or $B$ is true. Therefore if it was false then $A \land (A\rightarrow B)$ would be true and $B$ would be false.

If $A \land (A\rightarrow B)$ is false then A is false and $A\rightarrow B$ is false. But if $A\rightarrow B$ is false and $B$ is false then $A$ must be true, which is a contradiction.

6oncvlo
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To go from another direction, you can construct a proof using the deduction theorem to transform the problem first into $\{A \wedge (A \to B)\} \vdash B$. Then this is proved by the following:

  1. $A \wedge (A \to B)$ (hypothesis).
  2. $A$ ($\wedge$-elimination on 1).
  3. $A \to B$ ($\wedge$-elimination on 1).
  4. $B$ (modus ponens on 2 and 3).
Graham Kemp
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Patrick Stevens
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  • Please could you comment on your solution and explain in which sense there is no "circle in proving". The reason that makes it apparently circular is that modus ponens is used to prove modus ponens. Does the false appearence lie in a syntax/ semantics confusion or an object language / metalanguage one? – Floridus Floridi Feb 26 '21 at 22:55
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    Modus ponens is a rule of the metalanguage, but it's not automatically true of the theory. I've assumed we're working in a logic which includes modus ponens. Per https://math.stackexchange.com/questions/2376048/are-there-logics-without-modus-ponens, of course, there are logics which don't allow modus ponens, but I'm not aware of a logic without modus ponens in which you can prove something by two-valued truth table. – Patrick Stevens Feb 26 '21 at 23:01
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    To be precise, in the introductory course in propositional logic I took many years ago, a proof was defined as an ordered list of sentences, each either an axiom, a hypothesis, or a line of the form $q$ where an earlier line was of the form $p \to q$ and another earlier line was of the form $p$. In that setting, modus ponens was assumed as part of the background logic (it was the name given to that third rule), but it had no direct translation into symbols in a formal proof. – Patrick Stevens Feb 26 '21 at 23:03
  • Thanks for these comments. – Floridus Floridi Feb 26 '21 at 23:12
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A semantic reasoning ( not sure it counts as a " formal" proof)

Suppose the antecedent does not imply logically the consequent.

It means there is ( at least) one interpretation inwhich the antecedent is true and the consequent false.

Hence, an interpretation in which $B$ is false and $( (A\land (A\rightarrow B))$ is true ( meaning is particular that $A$ is true).

The only way to have $(A\rightarrow B)$ true is not to have both $A$ true and $B$ false.

But the only way to have this when you alredy have $B$ false is not to have $A$ true, hence to have $A$ false.

By what precedes,however, you have $A$ true.

This shows that there is no possible interpretation that falsifies the original material conditional, and, consequently, that the antecedent logically implies the consequent.

Floridus Floridi
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