I'm not talking about the truth values " true" / " false".   I'm talking about the symbols that are used in propositional calculs laws such as :   "False  OR  p is equivalent to p "  ( where "p" stands for any sentence).   I think that " truth" and " falsity" are called " constants".   I've already read things such as " False is the proposition that is equivalent to all antilogies".   But that does not seem satisfying, for any antilogy is equivalent to all antilogy, so one cannot say , I believe, " the proposition that is equivalent to any antilogy".   By the way, how did these symbols come to be introduced? What is their " history"?
121602
I have some doubts about the "natural" interpretation of $\bot$ in Natural Deduction and sequent calculus.  In Prawitz (1965) $\bot$ (falsehood or absurdity) is called a sentential constant [page 14]  Chiswell &amp; Hodges (2007) list $\bot$ (absurdity) between the truth function symbols [page 32], and this is not very clear for me; then, in the formal definition of formula of a language [page 33] they say :     $\bot$ is a formula.   Negri &amp; von Plato (2001) [page 2] list $\bot$ (falsity) between the prime formulas, specifying that :     Often $\bot$ is counted among the atomic formulas, but this will not work in proof theory. It is best viewed as a zero-place connective.   I think that the last comment is contra D.van Dalen, Logic and Structure (5th ed, 2013) [page 7] where $\bot$ is defined as a connective and :     The proposition symbols and $\bot$ stand for the indecomposable propositions, which we call atoms, or atomic propositions.   I'm wondering if all the above definitions are equivalent.  A (propositional) connective is an "operator" that maps one or more propositional variables into a formula; e.g.      $\land$ : &lt;$P,Q$> $\quad \rightarrow \quad P \land Q$.   This means that the zero-place connective $\bot$ is a mapping      $\bot$ : $\emptyset \quad \rightarrow \quad \bot$.   If so, may we say that, being at the same time the mapping and the output of the mapping, it is both a connective and a formula ?
94629
(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
57768
We could have written = as "equals", + as "plus", $\exists$ as "thereExists" and so on. Supplemented with some brackets everything would be just as precise.   $$\exists x,y,z,n \in \mathbb{N}: n&gt;2 \land x^n+y^n=z^n$$  could equally be written as:  ThereExists x,y,z,n from theNaturalNumbers suchThat       n isGreaterThan 2 and x toThePower n plus y toThePower n equals z toThePower n   What is the reason that we write these words as symbols (almost like a Chinese word system?)  Is it for brevity? Clarity? Can our visual system process it better?   Because not only do we have to learn the symbols, in order to understand it we have to say the real meaning in our heads.  If algebra and logic had been invented in Japan or China, might the symbols actually have just been the words themselves?  It almost seems like for each symbol there should be an equivalent word-phrase that it corresponds to that is accepted.
82466
Suppose we define the following relationship of ordering between propositions in PROP: $$ \phi &lt; \psi := (\models \phi \rightarrow \psi \land \lnot(\models \psi \rightarrow \phi) ) $$ Show that if exists $\phi$ and $\psi$ such that: $$ \phi &lt; \psi $$ Then there is a $\sigma$ such that: $$ \phi &lt; \sigma &lt; \psi $$ I have tried to use valuation to find what $\sigma$ could possibly be, so for all valuations v:  If $v(\phi)=0$ and $v(\psi)=0$ then, since $\models \sigma \rightarrow \psi$, $v(\sigma)=0$  If $v(\phi)=1$ and $v(\psi)=1$ then, since $\models \phi \rightarrow \sigma$, $v(\sigma)=1$  It is impossible that $v(\phi)=1$ and $v(\psi)=0$ since $\models \phi \rightarrow \psi$  If $v(\phi)=1$ and $v(\psi)=0$ then $v(\sigma)$ could be either 0 or 1, and it would satisfy the tautologies, but if:  $v(\sigma)=0$, then the valuation of $\sigma$ would be equal to valuation of $\psi$ for all valuations and thus $\models \psi \rightarrow \sigma$, which is a contraditcion; $v(\sigma)=1$, then the valuation of $\sigma$ would be equal to valuation of $\phi$ for all valuations and thus $\models \sigma \rightarrow \phi$, which is a contraditcion.
49575
I am currently doing the Intro to Mathematical Thinking course on Coursera. A quiz therein, asks:  For what truth values of Φ and Ψ do we define Φ ⇏ Ψ to be T (true)?  It says the right answer is Φ is true and Ψ is false.  I understand that; but I do not follow why the reverse is not right. That is, why does Φ false and Ψ true does not make the statement Φ ⇏ Ψ to be true.  I am confused because if Ψ is true while Φ is false, the truth of Ψ does not follow from the truth of Φ and so it should be that Φ ⇏ Ψ. Could someone please explain?
12286
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 ?
86376
If $\Gamma \cup \{ \neg \varphi \}$ is inconsistent, then $\Gamma \vdash \varphi$   Here, a set of formulas is inconsistent means they syntactically imply some formula as well as its negation. Syntactic implication here consists of (1) given formulas, (2) all first-order logical axioms, (3) modus ponens.  I have tried to explicitly construct a syntactic proof for the conclusion but failed to come up with one.
42531
We have the following argument: $P \implies Q$ (Premise 1) $P$ (Premise 2) $—$ $∴Q$ (Conclusion)  The accompanying truth table is:    I don't understand this section of my textbook:     The premises are both true only in line four of the table, and in this   line the conclusion is true as well.      You can also see from the truth table that both premises are needed to   make this argument valid. But if we were to change the truth table for   the conditional statement to make P → Q false in the first line of the   table, then the second premise of this argument would no longer be   needed.       We would end up with the conclusion that, just from the single   premise P → Q, we could infer that Q must be true, since in the two   lines of the truth table in which the premise P → Q would still be   true, lines two and four, the conclusion Q is true too.   My thinking:  If we would change the $P → Q$ to be false in the first line of the truth table. Then why couldn't the conclusion be false as well? I don't understand what line $2$ and $4$ have anything to do with line $1$?
64345
Verify is tautology by using logical equivalence:  $$((p ∨ q) ∧ (p → r) ∧ (∼ r)) → q$$  by showing what laws are using, step by step explanation needed.  Logical equivalence: de morgan's laws, commutativity, associativity, double negative, distribution, idempotent, absorption, identity, universal bound, negation  Rules of inference: modus ponens, modus tollens, generalisation, specialisation, conjunction, ellimination, transivity, proof by division into cases, contradiction  What I got so far:  I need to prove   $$((p ∨ q) ∧ (p → r) ∧ (∼ r)) → q ≡ \text{tautology}$$ Since $p → r$ same as $∼ p ∨ r$: $$((p ∨ q) ∧ (p → r) ∧ (∼ r)) → q ≡ ((p ∨ q) ∧ (∼ p ∨ r ) ∧ (∼ r)) → q$$  I am stuck here
5225
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.
116757
Most of the systems mathematicians are interested in are consistent, which means, by Gödel's incompleteness theorems, that there must be unprovable statements.  I've seen a simple natural language statement here and elsewhere that's supposed to illustrate this: "I am not a provable statement." which leads to a paradox if false and logical disconnect if true (i.e. logic doesn't work to prove it by definition). Like this answer explains: .  The natural language statement is simple enough for people to get why there's a problem here. But Gödel's incompleteness theorems show that similar statements exist within mathematical systems.  My question then is, are there a simple unprovable statements, that would seem intuitively true to the layperson, or is intuitively unprovable, to illustrate the same concept in, say, integer arithmetic or algebra?  My understanding is that the continuum hypothesis is an example of an unprovable statement in Zermelo-Fraenkel set theory, but that's not really simple or intuitive.  Can someone give a good example you can point to and say "That's what Gödel's incompleteness theorems are talking about"? Or is this just something that is fundamentally hard to show mathematically?  Update: There are some fantastic answers here that are certainly accessible. It will be difficult to pick a "right" one.  Originally I was hoping for something a high school student could understand, without having to explain axiomatic set theory, or Peano Arithmetic, or countable versus uncountable, or non-euclidean geometry. But the impression I am getting is that in a sufficiently well developed mathematical system, mathematician's have plumbed the depths of it to the point where potentially unprovable statements either remain as conjecture and are therefore hard to grasp by nature (because very smart people are stumped by them), or, once shown to be unprovable, become axiomatic in some new system or branch of systems.
93187
I have just started studying implication in mathematics. So, you'll probably have an idea of where my confusion lies right from the get go.  In the truth table, where $A \implies B$ we obtain this result:  A   B   T/F ___________ T | T | T T | F | F F | T | T F | F | T   Now, my confusion here is regarding why $A\implies B$ given A is false, and B true. The others I understand. The first and last one are obvious, the second one implies, to me anyway, that given A implies B, the truth of B rests upon the truth of A, B is false, A is True, which cannot be, thus not B given A is false.  Now... then, why is B true despite the fact that A is false? Or rather, why is the statement B given not A, True. The truth of B is implied from the truth of A, but A is a false statement. So it's very contradictory that this may be, and in fact, if you go through a real life example, you would find that the third statement seems to be false.  Say $A$ is the statement that it is dark outside, and $B$ the statement that the sun is not on our side of the earth. If it is not dark outside, than stating that the sun is on the other side of the planet must be wrong (in my example i'm going to claim that the sun is the only light source).  Side note:  I just had a thought, if A is false, does that mean you can conclude... mmm, whatever? I mean, B is implied via the truth of A, if A is false, B is NOT implied, but then B could be EITHER true or false, I suppose. Is this correct?  However, would that not mean that A does not imply anything when false. Would that not set the 3rd and 4th statement to be undefined?
46343
I am looking for a way to prove that the statement, $[(p \to q) \land (q \to r)] \to (p \to r)$, is a tautology without the help of the truth table. By using only Laws and Theorems like De Morgan's Law, Domination Law, etc. Also, I can't use the rules of inference. Please help, thank you.
29661
Why is "$P \Rightarrow Q$" equivalent to "$\neg Q \Rightarrow \neg P$"?  I am trying to understand why this does always apply, in terms of pure logic. Can you please explain it to me?
15976
For a proposition, such as:     P: Socrates is a man   Then not P is also a proposition:     Not P: Socrates is not a man   But do the same goes for properties? One can argue in the following way:    red is a property of red flags, or of red apples; but not red is not a property; for no thing instantiates not redness; that is not redness does not cohere in some substrate - here a flag, or apple; even though whiteness, which is an example of not redness, coheres in a white wall.  Q. Is this right?  All this, I think, takes it for granted that properties cohere in some substrate or substance - this is a commitment to a specific ontology; we may choose something else; the obvious choice being a world where all that there is are bundles of properties, and no substrate or substance to which to hang them on; there, existence is also a property - a property which all existing entities have.   Q. Given this ontological commitment; then is not P also a property?
87027
When we talk about formal logic, we generally agree that $P$ and $Q$ are deductively valid propositions. Below fact is supported by  Rutger Uni. page no.2 (PDF doc) And further when we talk about implication or material/concrete conditional $P\to Q$ we assume: &quot;if $P$ is true, then $Q$ is also true&quot; or more generally in formal logic we assume: if premises are true then conclusions are also true because we are dealing with deductively valid propositions. But when we draw truth-table of $P\to Q$ we come across one paradox. That is,  When $P$ is false, and $Q$ is true, $P\to Q$ is still true!  Why so? How can premise (deductively valid proposition or antecedent) be false  with a conclusion (deductively valid proposition or consequent) be true, and their implication be still TRUE! Reading through the explanations given on the question:  It is bit clear with the given examples that: When $P$ is false, and $Q$ is true, $P\to Q$ is true in some situations. Out of given examples I am selecting example given by Jai [3rd top answer on that page] which in fact comes from this :  &quot;If you get an A, then I'll give you a dollar.&quot; What if it's false that you get an A? Whether or not I give you a dollar, I haven't broken my promise.  Does this mean that if I get A+ or B- or I was ill for the exam, I will still get a dollar as promised? Because, I thought propositions are rigid and their use in conditional should be consider as-is without changing the meaning of them, then what is intended by the author of the given example. Here, it is very clear that author of the example suggest that; it is a necessary condition for &quot;Get an A&quot; is &quot;I'll give you a dollar&quot; and a sufficient condition for &quot;I'll give you a dollar&quot; is &quot;Get an A&quot;. Is there any further explanation: The premise (antecedent) requires further support or can be overlooked or ignored in order for conclusion (consequent) to be true when antecedent is false and consequent is true?
116659
In a  I am reading, the section on Propositional Logic says that a proposition is a statement that is either true or false, but not both true and false. Also, from  lecture online, the instructor says that we must be able to associate a truth value to a proposition.    The text I mentioned contains as an example of an assertion that is not a proposition the following:    (1) "this statement is false."    In the margin, the text says that the form of this statement makes it impossible to designate a truth value to it and the instructor in the lecture says simply that, "if [the statement] is true, then it is false, and if it is false, then it is true."    However, why exactly is it impossible to for (1) to have a truth value? What does it mean to say that if (1) is true, it is false, and conversely?    Response to Asaf Karagila As has been pointed out, I have already asked this question very recently  but it has not received proper attention. This question is one that I feel can be put to rest if only someone would provide an explanation that is direct and suitable for my level, which is that of a novice.