I'm completing the Stanford's "Introduction to Logic" course, and I'm having issues with the following exercise:  "Consider a language with object constants a and b, binary function constant c, and unary relation constants m and p and q. The definitions for the relations are shown below. Relation m is true of a and only a. Relation p is true of a structured object if and only if it is a linear list (as defined in Chapter 7) with a top-level element that satisfies m. Relation q is true of a structured object if and only if there is an element anywhere in the structure that satisfies m.  Your job is to show that any object that satisfies p also satisfies q. Starting with the axioms, use Fitch with Structural Induction to prove ∀x.(p(x) ⇒ q(x)). Beware: The proof requires more than 50 steps (including the premises)."  I think I'm getting there, but I've been stuck in step 24 for days now.    How can I go on? Thanks
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"Consider a language with a single object constant a, a single unary function constant s, and two unary relation constants p and q. We start with the premises shown below. We know that p is true of s(a) and only s(a). We know that q is also true of s(a), but we do not know whether it is true of anything else.  ¬p(a) p(s(a)) ∀x:¬p(s(s(x))) q(s(a)) Prove ∀x.(p(x) ⇒ q(x)). Hint: Break the problem into two parts - first prove the result for s(x), and then use that intermediate conclusion to prove the overall result."  I know I have to reach AX:(p(s(X)) => q(s(X))) and p(a) => q(a) first, but I'm struggling getting started. I also don't know if my first steps are correct. Any help? Thank you.
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This proof seems odd to me. I have come to the conclusion that I will use induction. I would like to see a smoother way or just some improvements on my technique.     Let $f:A\rightarrow B$ be a function between two finite sets of equal cardinality. Show that $f$ is surjective if and only if it  is injective.   To start, I will show that a surjection implies an injection using induction. I will dismiss the cases that both sets are empty or contain one element as being trivial (essentially vacuously true).   Assume $|A| \geq 2$, $|B| \geq 2$, $|A| = n = |B|$, and $f:A \rightarrow B$ is a surjection. For the base case, let $n = 2$.   There are two elements in both $A$ and $B$. Due to surjection, every element $b \in B$ must be mapped to, through $f$, by at least one element $a \in A$. If each of the two elements in $B$ were mapped to by the same element in $A$, the definition of function would be violated. Therefore, they are mapped to by unique elements in $A$. Thus, for $f(p), f(q) \in B$, if $f(p) = f(q)$, it must be true that $p = q$ so $f$ is injective.  Now assume that the surjection implies an injection for $n \geq 2$. We must show this to be true for $|A| = n + 1 = |B|$. Since it is true for $|A| = n = |B|$, the $n + 1$ case represents the addition of one new element to both $A$ and $B$. The new element in $B$ cannot be mapped to any other element in $A$ except for the new one. If mapped to by an old one, the definition of function would be violated. It must be mapped to by something since $f$ is surjective, hence it must be the new element. Finally, the new element in $A$ cannot be mapped to an old element in $B$ because it is unique and the previous $B$ was shown to be injective.  $$\blacksquare$$  This is a very wordy and awkward proof in my opinion. I have been out of proofs for a long time. I would like to see one that is more clear or seek validation if there isn't. I know that I have only completed half of the proof and have yet to go the other way.
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Is this proof correct?  p → q, ¬p → r, ¬q → ¬r ⊢ q  1) p → q      premise 2) ¬p → r     premise 3) ¬q → ¬r    premise   4) -q       assumption   5) -p       MT 1,4     6) p      assumption     7) ⊥      ¬e 5,6   8) ⊥        ¬e 2,3 9) q          ¬e 8   Am I going correct? What might be the next step?  I updated my answer.
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I have to prove if the following statement is true or false $$\forall x : (P(x) \lor Q(x)) \Leftrightarrow \forall x : P(x) \lor \forall x : Q(x)$$  I understand the first statement as, for every $x$, at least one of the functions $P$, $Q$ is true. For the second statement, at least one of the functions $P$, $Q$ is true for every $x$. Is this correct? When yes, how could I prove this? Using the distributive law will clearly render me the wrong result.
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Theorem Let be $(X,\le)$ a totally ordered set: then for any not empty finite subset $A$ of $X$ there exist the maximum  element and minimum element. proof. Let be $(X,\le)$ a totally ordered set and we prove by induction that any not empty finite subset $A$ of $X$ has a minimum element. Since $X$ is a totally ordered set, previously we observe that any its subset $Y$ (finite or infinite) is a chain. Obviously any subset $A$ of one element $a$ has trivially a minimum. So we suppose that any subset of $n$ elements has a minimum element and then  we consider a subset $A$ of $n+1$ elements: since $A$ is finite there exists a bijection $\phi$ from $A$ onto some natural number $m$, that is the successor of $n$, and so we can organize the elements of $A$  in a finite succession, that is $A=\{a_1,...,a_{n+1}\}$. Now we consider the subset $B=\{a_h\in A:h\le n\}$: obviously $X$ is a subset of $A$ that has $n$ element and so it has a minimum element $b$; so since $A=B\cup\{a_{n+1}\}$ and since $A$ is a chain (remember what before we observed), it must be or $a_{n+1}\le b$ or $b&lt;a_{n+1}$ and so for the transitivity property of the order relation $\le$ in any case $A$ has a minimum element. So now we only have to prove that any not empty finite subset $A$ of $X$ has a maximum element. So we consider the inverse relation $\preccurlyeq$ defined as $x\preccurlyeq y\iff y\le x$ for any $x,y\in X$: clearly $\preccurlyeq$ is a total order, since indeed $\le$ is a total order, and any minimum in $\preccurlyeq$ is a maximum in $\le$ and so since any not empty finite subset $A$ has a minimum in $\preccurlyeq$ it follows that any not empty finite subset in $\le$ has a maximum element. So we concluded the proof.  Is my proof correct? If not how prove the theorem? Could someone help me, please?
2322
I am trying to understand what mathematics is really built up of. I thought mathematical logic was the foundation of everything. But from reading a book in mathematical logic, they use "="(equals-sign), functions and relations.  Now is the "=" taken as undefined? I have seen it been defined in terms of the identity relation.  But in order to talk about functions and relations you need set theory.  However,  seems to be a part of mathematical logic.  Does this mean that (naive) set theory comes before sentential and predicate logic? Is (naive)set-theory at the absolute bottom, where we can define relations and functions and the eqality relation. And then comes sentential logic, and then predicate logic?  I am a little confused because when I took an introductory course, we had a little logic before set-theory. But now I see in another book on introduction to proofs that set-theory is in a chapter before logic. So what is at the bottom/start of mathematics, logic or set theory?, or is it circular at the bottom?  Can this be how it is at the bottom?  naive set-theory $\rightarrow$ sentential logic $\rightarrow $ predicate logic $\rightarrow$ axiomatic set-theory(ZFC) $\rightarrow$ mathematics  (But the problem with this explanation is that it seems that some naive-set theory proofs use logic...)  (The arrows are of course not "logical" arrows.)  simple explanation of the problem:  a book on logic uses at the start: functions, relations, sets, ordered pairs, "="  a book on set theory uses at the start: logical deductions like this: "$B \subseteq A$", means every element in B is in A, so if $C \subseteq B, B \subseteq A$, a proof can be "since every element in C is in B, and every element in B is in A, every element of C is in A: $C \subseteq A$". But this is first order logic? ($(c \rightarrow b \wedge b \rightarrow a)\rightarrow (c\rightarrow a)$).  Hence, both started from each other?
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Given $\exists y. \forall x. p(x,y)$, use Fitch-style natural deduction system to prove $\forall x.\exists y.p(x,y)$.  I know this question has been asked , but based on that answer I'm not able to move forward from step 3 below. I'm using  as a proof assistant by the way, maybe that's the issue. Any help is very much appreciated.  1.  EY:AX:p(X,Y)     Premise 2.  AX:p(X,Y).       Assumption 3.  p(X,Y).          Universal Elimination: 2
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Let $M,N$ be sets, $Q_M$ a set that depends on $M$ and $P(\cdot,\cdot)$ a property with two parameters (is it ok, to call these "parameters" or would "variables" have been better ?). Consider the following three statements:    i) $(\forall x\in M)(\forall y\in Q_M)[P(x,y)\Rightarrow x \in N]$   ii) $(\forall x\in M)[((\exists y\in Q_M)[P(x,y)])\Rightarrow x \in N]$    iii) $N=\{ x\in M \mid (\exists y\in Q_M)[P(x,y)]\}$  I would very much like to know the following:   I have been told that i) and ii) are equivalent, since they are an instance of a general scheme $$ (\forall x)(\Psi \Rightarrow \Phi) \Leftrightarrow  ((\exists z)\Psi) \Rightarrow \Phi),  $$ which holds if $z$ is not mentioned in $\Phi$, as Carl Mummert I think explained a long time ago  - I can't find the entry right now. What are the "values" the $x,z,\Psi$ and $\Phi$ have to take, such that this scheme shows the equivalence of i) and ii) ?      What kind of setting do you need in which you can prove this general scheme ? Is it only predicate logic ? Or do you need ZFC to formalize this logic so you can you use set-theory to prove it ?        How to you prove it (or a corrected version of it, if the above is flawed; it seems to me that quantifiers are missing) ? Is ii) equivalent to iii) ? How do you rigorously prove that ? Which ZFC axioms are needed for that proof ?
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Can someone give me an example for this question on reflexivity please. How is this reflexive? Can you show an example with numbers from the set?     Question 6   Let $X = \{0,1,2,3,\dotsc, 9\}$. Define a relation $\mathcal R$ from $X$ by "$x$ is related to $y$ if $x$ and $y$ gives the same remainder on division by $3$".   (a) Show that $\mathcal R$ is an equivalence relation.   Anwser:   (a) $\mathcal R$ is clearly both reflexive and symmetric, for every number has the same (unique) remainder on division by $3$ as itself and if $a$ has the same remainder on division by $3$ as does $b$, then $b$ has the same remainder on division by $3$ as $a$ also. Similarly $\mathcal R$ is transitive, for if $a\mathcal R b$ and $b\mathcal R c$, then $a$, $b$, and $c$ all have the same remainder on division by $3$, so $a\mathcal R c$ also.
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Problem statement:  Let A be an uncountable set, B a countable subset of A, and C the complement of B in A.  Prove that there exists a one-to-one correspondence between A and C.  My thoughts:  There's a bijection between A and A (the identity function).  There's a bijection between C and C (the identity function).  There's a bijection between B and $\mathbb{N}$.  That's all I know.
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Prove that structure $(Q/{0}; =;*)$ has infinite number of automorphisms I think it's about we don't have $&lt;$ or $&gt;$, so we don't have strict order in structure, so we can do whatever we want? but i dont know how to say it more formal/
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Let $r$ be a relation on $A \times A$ such that $(a,b) r (c,d) \iff ad = bc.$ How can I show that this relation is transitive, ie. $(a, b)r(c,d)$ and $(c,d)r(e, f) \implies (a,b)r(e,f)$?  I tried to say that $(a,b) r (c,d)$ means that $c=ka$ and $d=kb,$ $k$ a coefficient, such that $ad = bc \iff akb = bka,$ and going on from there, but I'm not sure this is valid for all values of $a, b, c,$ and $d$. How can I show the transitivity of the relation?
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I am wondering if there is a standard definition for a category with objects as first order logical (FOL) expressions e.g. $\neg x \vee y$. It seems to me that these logical expressions would be part of a concrete category as they can be mapped (forgetfully) to sets, and as many concrete categories (groups etc.) have homomorphisms as morphisms that perhaps homomorphisms are the morphisms in this FOL category.  I am having difficulties visualizing what these homomorphisms would look like - perhaps just some operation $ (\circ,\bullet) $ such that   $f((x \wedge y) \vee z) \rightarrow (f(x) \bullet f(y)) \circ f(z)$ where $(\vee,\wedge) \rightarrow (\circ,\bullet)$  ?  Thanks,  Brian
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$((\neg \alpha \to \alpha) \to \alpha) $  Hi, I am trying to prove this.  Can someone gives me some hints to start the question... My friend told me I might need to use deduction theorem here, but I have no clue to approach it..  Axiom 1: A→(B→A).  Axiom 2: (A→(B→C))→((A→B)→(A→C)).  Axiom 3: (¬B→¬A)→((¬B→A)→B).  To clarify: A, B, C, α, and β are propositions (i.e. assigned True or False). → and ¬ have the standard logical meanings.
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Two fallacious arguments:   If P, then Q  If P, then R  Therefore: If Q, then R   And   If P, then Q  If R, then Q  Therefore: If P, then R   However, if these particular propositions were interpreted as being connected not by a conditional sign but by an "=" (identity) sign, wouldn't we have examples of the transitive property (i.e. P=Q, P=R, so Q=R)?
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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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I've scoured the math stackexchange and the philosophy one for some guidance on how to go about this while using the Fitch System.  Anyone can attempt it here;    I can reach the conclusion, but it's not in the correct "layer" of the proof. I'm stuck with the answer as a subproof.  (Note that the fitch system lacks contradiction elimination (RAA) as a method for getting out of assumptions).