I am having trouble with this Natural Deduction question  $$(∀x∃y (P(x) → Q(y))) → (∃y∀x(P(x) → Q(y)))$$
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I have working on this formula $(\forall x \exists y (P(x) \supset Q(y))) \supset (\exists y \forall x (P (x) \supset Q(y)))$ to either prove or disprove it.  First, I tempted to disprove it, but I changed my mind.  I wrote down "for all x that there exists some y satisfies corresponding condition", and "there exists some y that for all x satisfies corresponding condition." I think these statements refer to the same idea.  Any suggestions?
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This came up in proving non-regularity of a certain language (powers of 2 over the ternary alphabet). Any clue to the above equation could help me move forward.  Edit: Of course, $x = 1, y = 1$ is a solution. I am looking for non-trivial solutions.
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I am reading about the proof of law of expectation for the discrete type,   $$ \begin{align} E[E[X|Y]] &amp;=  E \left[ \sum_{x} x \cdot P(X = x | Y) \right] \\ &amp;= \sum_y \left[ \sum_{x} x \cdot P(X = x | Y = y) \right] P(Y = y) \tag 2) \\ &amp;= \sum_x x \sum_y  P(X = x | Y = y) \cdot P(Y = y) \tag3) \\ &amp;= \sum_x x \sum_y P(X = x \, \text{and} \, Y = y) \tag4) \\ &amp;= \sum_x x \cdot P(X = x) \tag5)\\ &amp;= E[X] \end{align} $$     Question: Why does it assume Y=y? And can someone give me justification of each step?
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Just for clarity, this is the specific formulation I am referring to.    $K$: $\Box(p\to q)\to(\Box p\to\Box q)$  Naturally, we might consider the converse of this statement:    $K'$: $(\Box p\to\Box q)\to\Box(p\to q)$  The question is simple. Given $K$, does $K'$ follow? An accompanying proof of the answer would be greatly appreciated.
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Here is the proof:     But I do not understand the following:  1-why the function needed to be bilinear to use the universal property?  2- what is he doing starting from the paragraph that starts with the statement "In the other direction ...." ?
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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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Using only rules of natural deduction, I am trying to prove that $$p \implies q \vdash \lnot p \lor q$$ but am having a lot of difficulty. I was able to prove the other direction.   Could anyone show me or point me to the right direction?
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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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I know when evaluated it gives one but I can't figure out how to prove it.  Can anyone help? I believe it requires L'Hopitals rule and taking a natural log but i cannot figure out the exact math.
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Here's another of Tomassi's exercises I can't solve (Logic, page 106):  ~ (~P &amp; ~Q) : P ∨ Q   I have to use natural deduction and the only rules I know are:   assumptions, modus ponendo ponens, modus tollendo tollens, double negation, reductio ad absurdum, conditional proof, v-introduction, v-elimination, and introduction, and elimination.   Tomassi's proof consists of 14 steps.  Thanks for the help.
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Does the quantifier order matter in this problem? I don't think it does.  (∃y in R)(∀x in R)(x+y=x) For some real number y and for all real numbers x, x plus y equals x. I assert this to be true. (I won't bother with the proof unless somebody really wants it)  I feel that, although the quantifiers are switched, the sentence is still true. Is this right?  (∀x in R)(∃y in R)(x+y=x) For all real numbers x there is a real number y such that x plus y = x  I also feel that these sentences are also all true, regardless of their order. Am I missing something here?   (∃y in R)(∀x in R)(x+y=0) and (∀x in R)(∃y in R)(x+y=0).   Plus these two sentences (∃y in R)(∀x in R)(xy=1) and (∀x in R)(∃y in R)(xy=1)
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I'm thinking of $f$ as a function either from the reals to the reals or from the naturals to the naturals.   Edit: Okay, what if $f$ is from the naturals to the naturals?
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I attempted a proof of mathematical induction using an arbitrary base case, but was unsuccessful (and hence this question). Below is what I was trying to do and along with my thinking; if anyone can point me in the right direction I'd appreciate it.  The induction I am using: Let $P$ be a property about the natural numbers $\mathbb{N}$ with $0\in\mathbb{N}$, and let $P(n)$ denote the statement that the property $P$ holds for $n\in\mathbb{N}$. Suppose $P(0)$. Furthermore suppose that for each natural number $k$, $P(k)$ implies $P(k+1)$. Then $\forall nP(n)$.  What I am trying to prove: Let $P$ be a property about the natural numbers $\mathbb{N}$ with $0\in\mathbb{N}$, and let $P(n)$ denote the statement that the property $P$ holds for $n\in\mathbb{N}$. Suppose for $n_0\in\mathbb{N}$, $P(n_0)$. Furthermore suppose that for each natural number $k\geq n_0$, $P(k)$ implies $P(k+1)$. Then $\forall n\geq n_0 P(n)$.  My attempted proof: define $Q(n)$ to be $n\geq n_0 \to P(n)$. Then we wish to prove $\forall nQ(n)$. We induct on $n$.  Base case (for ordinary induction): $Q(0)$ is $0\geq n_0\to P(0)$. Since $n_0\in\mathbb{N}$, $0\geq n_0$ implies that $n_0=0$. Since $P(n_0)$, $P(0)$, which proves the base case.  Inductive step: we want to show $\forall n(Q(n)\to Q(n+1))$. To do this, we assume $k\geq n_0 \to P(k)$ and try to show $k+1\geq n_0 \to P(k+1)$.  Since $k\geq n_0 \to P(k)$, we first prove the case where $k\geq n_0$ and $P(k)$. By the hypothesis of the proof, we see that $P(k+1)$, which proves the case for $k+1$.  This is where I am having trouble: For $k&lt;n_0$, I am unable to show the implication for $k+1$.  So my questions would be: (1) is the overall approach for the proof correct? (2) If so, how might I go on to prove the case when $k&lt;n_0$?  Thanks in advance. (This is not homework, by the way.)
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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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Let $x,y \in L_p$   such that  $\|x\|_p=\|y\|_p=1$ , $1&lt; p&lt;\infty$ and $x\neq y.$ Why is $\|x+y\|_p&lt;2$  ?  I'm not sure how to start the proof.. I don't know how to handle integral of $(x+y)^p$ and it seems that using the binomial theorem won't be a great success.
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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'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.