Can anybody please explain to me how "if $p$, then $q$" is logically equivalent to "$q$ unless ~$p$" ?  My problem is with all truth possibilities of $p$ and $q$ except for when both are true It seems in those situations they would not have the same logical meaning For e.g: "if $p$, then $q$" is true if $p$ is false and $q$ is true. However, shouldn't this possible combination render "$q$ unless ~$p$" false? Feel free to use truth tables or whatever method that would help in explanation.
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Why isn't a conditional statement said to "not apply" instead of be "vacuously true" if the hypothesis is not satisfied? That would seem more appropriate.
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I understand that  (1) "$A$ if and only if $B$" ($A\iff B)$  means that  (2) "$A$ implies $B$ and $B$ implies $A$" $(A\implies B)\land (B\implies A)$.  The phrase "$A$ if and only if $B$" sounds as though something applies exclusively or precisely in case something else is true, i.e., not only does $B$ make $A$ true, but $B$ is the only thing that makes $A$ true. If this is the case, then how exactly does this interpretation relate to two things implying one another?    For the Bounty: (1) I think that a natural interpretation of the statement $A$ if and only if $B$ is that $A$ is true precisely in case $B$ is true, i.e., if $B$ is true, then $A$ is true, and if $B$ is false, then $A$ is false. Is this correct? And if so, how does this interpretation mean that $A$ and $B$ imply one another and also have the same truth value?    (2) Related to (1), the (true) statement $\forall x\in\mathbb R:2x=5\iff x={5\over 2}$ means that for any replacement of the variable, the two statements have the same truth value. How does two statements having the same truth value in all cases mean that they imply each other?  (3) Most importantly, I need exposition. Am I missing anything obvious, or is this a simple thing?
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I know, I know, there are tons of questions on this -- I've read them all, it feels like. I don't understand why $(F \implies F) \equiv T$ and $(F \implies T) \equiv T$.  One of the best examples I saw was showing how if you start out with a false premise like $3=5$ then you can derive all sorts of statements that are true like $8=8$ but also false like $6=10$, hence $F \implies T$ is true but so is $F \implies F$.  But for me examples don't always do it for me because how do I know if the relationship always holds even outside the example? Sometimes examples aren't sufficiently generalized.  Sometimes people say "Well ($p \implies q$) is equivalent to $\lnot p \lor q$ so you can prove it that way!" except we arrived at that representation from the truth table in the first place from disjunctive normal form so the argument is circular and I don't find it convincing.  Sometimes people will use analogies like "Well assume we relabeled those two "vacuous cases" three other ways, $F/F, F/T, T/F$ -- see how the end results make no sense?" Sure but T/T makes no sense to me either so I don't see why this is a good argument. Just because the other three are silly doesn't tell me why T/T is not silly.   Other times I see "Well it's just defined that way because it's useful"... with no examples of how it's indeed useful and why we couldn't make do with some other definition. Then this leads to the inevitable counter-responders who insist it's not mere definition of convenience but a consequence of other rules in the system and so on, adding to the confusion.  So I'm hoping to skip all that: Is there some other way to show without a doubt that $(F \implies q) \equiv T$?
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This semester I take Introduction to logic and sets theory and for logic our reference is A Mathematical Introduction to Logic by H.B.Enderton.  I have some problems with this course I mean it is somehow unfamiliar ! our professor teach well and I learn at class but when I want to solve my homework ,I have no idea ,how to solve problems !  I don't have any idea how to prove this :  suppose a and b are tautologically equivalent, where A is wff and a is used in A and B is  another wff that b is used in; prove that A and B are tautologically equivalent.  Any one can help me ?
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Without using the truth table, I need to prove:  $$\big((p\rightarrow q) \lor (p \rightarrow r)\big) \rightarrow (q\lor r)\equiv p \lor r \lor q$$  Up until now, we've been using truth-tables to verify equivalences. So I'm a bit lost on where to begin without using a truth-table.
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Using the Fitch system, how do I prove $((p \implies q) \implies p)\implies p$? I started with the hypotheses $(p \implies q) \implies p$ and $\sim p$. However, from these hypotheses I did not get the desired contradiction to solve the question.  Thanks sb45
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(p → q) ∧ ¬p ⇒ ¬q  I have tried to use the rules of inference in many ways to prove this argument to be valid. However, I cannot seem to find a way to validate the argument, but I am afraid I have missed something.  Using a truth table I find that when p and q is 0, then p → q is 1, ¬p is 1 and ¬q is 1. Would that not make the argument valid?  I apologize in advance if I have misunderstood the whole concept. Any answer would be much appreciated. Thank you
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How is the statement $\bot \to \top$ true? If, for example, I take the statement  $P = 2$ is an odd number $Q = 5$ is an odd number Then $P$ is false and $Q$ is true.  Clearly, false implies true translates to  if $2$ is an odd number then 5 is an odd number  which means that  if $2$ is not an odd number (i.e. even) then $5$ is not an odd number  which is clearly false as $2$ is indeed an even number. Please help me to find what I’m missing out in this. Thanks for your help!
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Allowed inference rules: ∨-I, ∨-E, ∧-I, ∧-E, →-I, →-E, ¬-I, ¬-E  I tried to prove a contradiction by assuming $¬ ((P → Q) ∨ (Q → P))$ but got stuck, or am I doing it in the wrong way?  Edit:  My proof attempt  $1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad Assum$  $2.\qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  Assum$  $3.\qquad(P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\; ∨-I(2)$  $4.\qquad(¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\; ∧-I(1,3)$  $5.\qquad  ¬ (P → Q)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\; ¬-I(2,4)$  $6.\qquad  Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\quad\; Assum$  $7.\qquad  (P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\; ∨-I(6)$  $8.\qquad  (¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\;\; ∧-I(1,7)$  $9.\qquad  ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; ¬-I(6,8)$  $10.\qquad ¬ (P → Q) ∧ ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\;\qquad\qquad\quad\;\;\, ∧-I(5,9)$  Stuck at this line, can't spot any contradiction to line 1
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I am struggling with the following problem.  nb: this is all in propositional logic (P).    Show that the following are equivalent:  (1). $\Gamma $ is consistent and complete.  (2).  There is exactly one truth assignment that satisfies $ \Gamma $.   My weak attempt:  $\Rightarrow$ Direction:  As $ \Gamma $ is consistent, we have that $ \not \exists A, \text{formula}, st ~\Gamma \vdash A ~\text{and}~\Gamma \vdash \neg A $   As $ \Gamma $ is complete, we have that $ \forall A, \text{formula}, \Gamma \vdash A ~\text{or}~ \Gamma \vdash \neg A $  truthfully,I struggle with moving past these definitions! Especially getting from he syntax to the semantics of it all, in order to incorporate truth assignments.    Please Help!  thank you:)
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This question comes from from my algebra paper:  $(p \rightarrow q)$ is logically equivalent to ... (then four options are given).   The module states that the correct option is $(\sim p \lor q)$. That is:      $$(p\rightarrow q) \iff (\sim p \lor q )$$   but I could not understand this problem or the solution. Could anybody help me?
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Suppose A is a necessary condition for B. Does this mean $A\Rightarrow B$ or does it mean $B\Rightarrow A$?  Suppose P is a sufficient condition for Q. Does this mean $P\Rightarrow Q$ or  $Q\Rightarrow P$?
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Could someone show me the truth table for this proposition? I think I have the last two down, but I'm not sure what the symbol in the following one is: $$p\oplus (p\wedge q)$$
80488
I am currently learning about first order logic, and as an exercise was asked to provide free and bound variables in the following formula:  $$(\exists x P(y,y) \implies \exists y P(y,z))$$  I'm currently confused on the following: the term $P(y,y)$ clearly has $y$ as a free variable, while $\exists y P(y,z)$ clearly has it as a bound variable.  From definition, it should be that in the formula above, $y, x$ are both bounded and $z$ is free. However, this does not convey that if I were to do a "change of variables" to $(\exists x P(k, k) \implies \exists y P(y,z))$, then $x, y$ would be bound and $k, z$ would be free, which means that the number of free variables actually changed for formulas that are actually "equivalent" (since both $P(y,y)$ and $P(k,k)$ are referring to a constant, rather than a variable).  I am also having trouble understanding what the full parenthetization of this expression would be: $(\exists x (P(y,y) \implies (\exists y P(y,z))))$ or $((\exists x P(y,y)) \implies (\exists y P(y,z)))$. Both would a priori be valid, and even though it doesn't really change anything in this case (since $x$ is not mentioned within the formula that is being quantified), I'm curious.
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I am studying entailment in classical first-order logic.  The Truth Table we have been presented with for the statement $(p \Rightarrow q)\;$ (a.k.a. '$p$ implies $q$') is: $$\begin{array}{|c|c|c|} \hline p&amp;q&amp;p\Rightarrow q\\ \hline T&amp;T&amp;T\\ T&amp;F&amp;F\\ F&amp;T&amp;T\\ F&amp;F&amp;T\\\hline \end{array}$$  I 'get' lines 1, 2, and 3, but I do not understand line 4.      Why is the statement $(p \Rightarrow q)$ True if both p and q are False?  We have also been told that $(p \Rightarrow q)$ is logically equivalent to $(~p || q)$ (that is $\lnot p \lor q$).  Stemming from my lack of understanding of line 4 of the Truth Table, I do not understand why this equivalence is accurate.       Administrative note. You may experience being directed here even though your question was actually about line 3 of the truth table instead. In that case, see the companion question  And even if your original worry was about line 4, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.
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For some reason, be it some bad habit or something else, I can not understand why the statement "p only if q" would translate into p implies q. For instance, I have the statement "Samir will attend the party only if Kanti will be there." The way I interpret this is, "It is true that Samir will attend the party only if it is true that Kanti will be at the party;" which, in my mind, becomes "If Kanti will be at the party, then Samir will be there."   Can someone convince me of the right way?  EDIT:  I have read them carefully, and probably have done so for over a year. I understand what sufficient conditions and necessary conditions are. I understand the conditional relationship in almost all of its forms, except the form "q only if p." What I do not understand is, why is p the necessary condition and q the sufficient condition. I am not asking, what are the sufficient and necessary conditions, rather, I am asking why.
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I hope to show that the comprehension axioms follows from the replacement schema.  This is a solution that professor wrote.  $P(u,u)$: every set $u$, exists an unique $u$ such that $\psi(u)$.  Then exist a set $y=\{u:P(u,u) \text{ and } u \in X\}$  Let $F=\{(u,u) : \psi(u)\}$. Then $F(X) \in u$ and $y=\{u \in F(x) : \psi(u)\}$  I took some notes but I think that there are much errors.  Thanks in advance.