How can we give a meaning to the three axioms of prepositional logic? As axioms, is not it supposed to be obvious? For example, how is $(\phi \to (\psi \to \chi)) \to ((\phi \to \psi) \to (\phi \to \chi))$ evident?
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How might I go about getting some intuition on the typical axiom schemes given for propositional logic? They seem rather mysterious at first glance.  For example, these are taken from:    $\phi \to \phi$ $\phi \to \left( \psi \to \phi \right)$ $\left ( \phi \to ( \psi \rightarrow \xi \right)) \to \left( \left( \phi \to \psi \right) \to  \left( \phi \to \xi \right) \right)$ $\left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right)$
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I have a feeling it's not, because ¬¬P → P is not provable. If it is, I'm not sure what kind of reductio I'd need to negate ¬(¬¬P → P). I believe a textbook somewhere said it was provable in intuitionistic logic, so am I missing something or is the textbook wrong?
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I am really lost trying to understand the difference between the logical connective "implies", $\to$, and the metalogical symbol (or maybe it's also a connective?) $\vdash$. (This is all focusing on prepositional logic here).  In metalogical terms, for example with modus ponens, it is said that $P, P \to Q \vdash Q$, which means "If we have a proof of $P$, and we have a proof of $P \to Q$, then we can infer / make a proof of $Q$". But I don't understand what the difference is between that and saying something like $P \land (P \to Q) \to Q$ which is similar but uses $\to$ instead of $\vdash$.  For example the $P \to Q$, at least in my experience, means "it is possible to go from $P$ to $Q$" but I don't see how "going to $Q$" is different from "inferring $Q$." Simply telling me that one is metalogical and one is not doesn't really help me understand what's going on.   I've also been given the example of  but I don't understand this, either. It sounds like the tortoise is constantly rejecting implications because "who says I have to accept conclusions just because the premises are true?" but then somehow introducing a metalogical $\vdash$ solves this? "We use the metalogical symbol $\vdash$ to basically force that stubborn tortoise to accept the conclusions and we've now circumvented the issue."  Unless I've grossly misunderstood something I just don't see why that's even a thing. Who says then I have to accept $\vdash$? Is $\vdash$ just a stronger form of $\to$, like a "sudo $\to$" or something (to borrow a Linux term), a form of implies that forces the conclusion to be accepted from the premise(s)?   What's the difference? How are they working here? Why do we need them? Are there any concrete examples showing the difference of both?
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I'm having a bit of trouble proving this with the propositional calculus rules.  $$p \lor (p\implies q)$$ Would someone mind helping me and showing which rules they've used with an explanation! Thanks!
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I studied propositional logic, and everyday I see applications of what I learned on the internet, in mathematical books and miscelaneous resources.        One particular case is sentences in the form $a \rightarrow b$ , that is, material implications. While I fully understand the mathematical logic approach to the study of those kinds of formulas, sometimes I get confused about what someone ( using implication sentences in the ordinary language ) is trying to say.        In ordinary language, I often see affirmations in the form of an implication.  One example would be someone affirming that "the moon is made of cheese" $\rightarrow 2+2=4$. [ Kleene, Mathematical Logic ].      I perfectly understand that if we choose the interpretation that gives the antecedent as false ( and the consequent as true ) we get that the whole implication is also true, but I would like to understand EXACTLY what he is trying to say when he AFFIRMS a sentence in the form of an implication. Is he saying that under the commonly agreed interpretation of that sentence ( antecedent = false, consequent = true ), that implication becomes true?      Another example, totally different, would be someone affirming that $x \in \mathbb{N} \rightarrow x \in \mathbb{N}$ or $x \in \mathbb{Z}_{&lt;0}$. Is he now affirming the same kind of sentence or saying something totally different?      When he affirms that sentence, is he saying that he finds the sentence to be a tautology?   Why didn't he say that?          Then, with these two examples we have an affirmation of the same kind of sentence (implication) meaning completely different things, in one case he affirms to say that under the commonly agreed interpretation of the sentence, it is true. In another case, he affirms to say that there's no interpretation that makes the sentence false.       Am I missing something? There is really that ambiguity in ordinary language, when people are affirming sentences?  Can't I just take some general instance of the affirmation of a sentence in the form $a \rightarrow b$ in ordinary language and know exactly what he's trying to say, regardless of the propositions he chooses the substitute $a$ and $b$ for?       Thanks a bunch.
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Every text I've come across uses the Axiom of Extensionality to describe the fact that sets are equal iff they contain each other's elements. $$(\forall x)(\forall y)\bigl((\forall a)(a\in x \iff a\in y) \implies (x=y)\bigr)$$ Some will regard this as an axiom, while some regard this as the very definition of equality (replacing the implication with a biconditional). Either way, the idea is that given two sets, regardless of how they were named, defined, or written, if they have exactly same elements, they are equal. A set can be described completely and only by its elements.  Another fun fact about sets is that equal sets are members of exactly the same collections. $$(\forall x)(\forall y)\bigl((x=y) \implies (\forall b)(x\in b \iff y\in b)\bigr)$$ I've always been comfortable with the idea that equal sets have the same elements, but the second fact is, to me at least, much more fascinating, and strikes a beautiful symmetry with the first.  If we choose to call the first fact the "Axiom of Extensionality" (which seems to be the convention), what is the name of the second fact? Is it another axiom, or can it be proven from the first? Are they logically equivalent? If we take Extensionality as a definition, then the second fact reduces to $$(\forall x)(\forall y)\bigl((\forall a)(a\in x \iff a\in y) \implies (\forall b)(x\in b \iff y\in b)\bigr)$$  Are there any texts that mention this? So far I have not seen it other than on .
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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.
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Okay I have the following fitch proof almost figured out, from here I know I have to move forth in conlcuding but I don't know what I am missing  (2 premises)  S → (R ∨ P)  P → (¬R → Q)   my work below  new scope assume S   Rule: > elim  R ∨ P  New scope assume R  Q∨R     Rule: V intro  End scope – (new scope assume P)  ¬R → Q         Rule: > Elim  my confusion comes where I go from here  Q∨R            rule Taut Con ??  Goal:├ S → (Q ∨ R)
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I've been doing some practice questions in the textbook for an upcoming predicate calculus lecture and I think I've managed to get A and B (possibly C) correct but I am clueless on how I can manage D and E.   For A I got: ∃x(R(x)∧¬B)  For B I got: ∀x(¬R(x)∧B)  For C I got: ∃x(R(x)∧(x=y))   - Not sure if this is even remotely correct    R(X) : ”X is rich”  X = Y : ”X equals Y”  B : ”Bill Gates”  a) Someone besides Bill Gates is rich.  b) Bill Gates alone is rich.  c) At least two people are rich.  d) Exactly one person is rich.  e) Exactly two persons are rich.  The universe of discourse is the set of all people.
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It's all in the question really. I am working on a proof in Fitch for a class, but I am very much stuck.  I am proving the tautology that "(P → Q) ↔ (¬P ∨ Q)", and I have already finished half of it, but now I must prove that "(¬P ∨ Q)" implies "(P → Q)". I can't seem to get anywhere.   I try to set up a proof by cases where I assume in different subproofs "¬P" and (in the other) "Q", but then I must prove "P → Q" from those. It seems even more difficult. Any help would be appreciated.
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In Chiswell and Hodges Mathematical Logic the authors define a sequent as such      "A sequent is an expression   (Γ ⊢ ψ) (or Γ ⊢ ψ when there is no ambiguity)   where ψ is a statement (the conclusion of the sequent) and Γ is a set of statements (the assumptions of the sequent) ...There is a proof whose conclusion is ψ and whose undischarged assumptions are all in the set Γ.".   They then go on to provide the axiom as follows:     Sequent Rule (Axiom Rule) If ψ ∈ Γ then the sequent (Γ ⊢ ψ) is correct.   Upon further reading about what exactly it means to be a sequent  I understand that for $\psi$ to be satisfied or "correct", every element of   $\Gamma$ must be true as their all linked by AND conjunctions.    What confuses me is that suppose that there is an element in $\Gamma$ which is false then the whole antecedent is false according to the wikipedia definition by conjunction and then even though $\psi$ exists and is true in $\Gamma$ ultimately $\psi$ would be false. Could someone please explain to me and help clarify how the axiom and the definition can both be true. Thank you for all of your help.
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I have no idea whether this question is way too specific or whether something similar has already been asked (we still need to work out a way to search for formulas I guess).  Anyways here I go: I want to show that some choice of (logical/quantifier) axioms for a Hilbert derivation system (first order logic, one sort) can prove all the axioms of some other choice.  I have the following at my disposal:  Axioms:   (fot): first order tautologies (i.e. inserting formulas for the propositions in a propositional tautology) (aax): $\forall x A[x] \to A[t]$ (where $t$ is any term and we assume no conflicts with free variables getting bound, $[...]$ denotes replacement)   Inference rules:   (mp): modus ponens (aru): $\frac{A\to B}{A\to\forall x B}$ where $x$ is not free in $A$.   I want to show that this system proves the formula:  $\forall x (A\to B) \to(A \to \forall x B)$ for $x$ not free in $A$ (which I will assume in the following).  This is of course very similar to (aru), but I want this formula, which is of course a stronger statement. I am quite lost here.  Here are my failed attempts: With (aax) and (aru), I easily get the provable rule $\frac{\forall x(A\to B)}{A\to\forall x B}$ which is the formulation of my goal formula as a rule (weaker).  I even get   $$\frac{\bar\forall(A\to B)}{A\to\forall x B}$$ where $\bar\forall$ denotes the universal closure of the following formula.  Using the Deduction Theorem (which I proved for this theory) I get that my system proves $$\bar\forall(A\to B)\to(A\to\forall x B).$$ but that is not exactly what I am looking for.
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This is all in the context of a Hilbert system with modus ponens $A, A\to B \vdash B$ and axioms:  Axiom $1$: $A \to (B \to A)$   Axiom $2$: $(A \to (B \to C)) \to ((A \to B) \to (A \to C))$   Axiom $3$: $(\lnot B \to \lnot A) \to (A \to B)$  So I tried to prove double-negation elimination by doing the following:  \begin{array}{ccc} 1 &amp; \{ \lnot \lnot a \} \vdash \lnot \lnot a &amp; \text{Premise}\\ 2 &amp; \{ \lnot \lnot a \} \vdash \lnot \lnot a \to (\lnot \lnot \lnot \lnot a \to \lnot \lnot a)  &amp; \text{ Axiom } 1\\ 3 &amp; \{ \lnot \lnot a \} \vdash \lnot \lnot \lnot \lnot a \to \lnot \lnot a  &amp; \text{ Modus Ponens } 1, 2\\ 4 &amp; \{ \lnot \lnot a \} \vdash (\lnot \lnot \lnot \lnot a \to \lnot \lnot a) \to (\lnot a \to \lnot \lnot \lnot a)  &amp; \text{ Axiom } 3\\ 5 &amp; \{ \lnot \lnot a \} \vdash \lnot a \to \lnot \lnot \lnot a  &amp; \text{ Modus Ponens } 3, 4\\ 6 &amp; \{ \lnot \lnot a \} \vdash (\lnot a \to \lnot \lnot \lnot a) \to (\lnot \lnot a \to  a)  &amp; \text{ Axiom } 3\\ 7 &amp; \{ \lnot \lnot a \} \vdash \lnot \lnot a \to  a  &amp; \text{ Modus Ponens } 5, 6\\ 8 &amp; \{ \lnot \lnot a \} \vdash a  &amp; \text{ Modus Ponens } 1, 7\\ 9 &amp; \vdash \lnot \lnot a \to a  &amp; \text{ Deduction Theorem } \\ \end{array}  And then I continued on, to prove double negation introduction:  \begin{array}{ccc} 10 &amp; \vdash \lnot \lnot \lnot a \to \lnot a &amp; \text{Double Negation Elimination}\\ 11 &amp; \vdash (\lnot\lnot \lnot a \to \lnot a) \to (a \to \lnot \lnot a) &amp; \text{ Axiom } 3\\ 12 &amp; \vdash a \to \lnot \lnot a &amp; \text{ Modus Ponens } 10, 11\\ \end{array}  My questions:   Am I allowed to invoke the deduction theorem in the middle of a proof like this (step $9$)? Is this considered a context change? Is context-changing allowed? Are the proofs even right/correct? Assuming it's all correct: Are we able to combine them to say that $a \to \lnot \lnot a$ and $\lnot \lnot a \to a$, therefore $a \iff \lnot \lnot a$ or is this symbol not allowed? What ultimately let us say $\lnot \lnot a = a$, the use of the equal sign here?
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In fact I don't understand the meaning of the word "metamathematics". I just want to know, for example, why can we use mathematical induction in the proof of logical theorems, like The Deduction Theorem, or even some more fundamental proposition like "every formula has equal numbers of left and right brackets"?  What exactly can we use when talking about metamathematics? If induction is OK, then how about axiom of choice/determincacy? Can I use axiom of choice on collection of sets of formulas?(Of course it may be meaningless. By the way I don't understand why we can talk about a "set" of formulas either)  I have asked one of my classmates about these, and he told me he had stopped thinking about this kind of stuff. I feel like giving up too......
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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.
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I'm working with the universal and the existential quantifiers and I was wondering is there an order of operations that applies to them? What is the difference between   $(\forall x)P(x)(\exists x)Q(x)$  And  $(\exists x)Q(x)(\forall x)P(x)$  Lets say we have this relationship  (a) $[(\forall x)P(x) \vee (\exists x)Q(x)] \implies (\forall x)[P(x) \vee Q(x) ]$
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(1)$\exists x \in X \ ( P(x)  \implies \forall y \in X P(y) )$  (2) $\exists x\in X P(x) \implies \forall y \in X P(y)$  What’s the difference? And are they both always true?   It seems to me that the second is false, but I read that the first is true. May someone please clarify the difference? And tell me why that is the case, please?