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I'm not sure what your example has to do with the Incompleteness Theorem. If the Goldbach Conjecture is false, then yes, your program will eventually find a counterexample and thus prove it to be false. But if it is true, we might still be able to prove it true as well. For example, suppose I have a computer program that tests the hypothesis that every even ...


0

(Posted after a previous answer was accepted) Using a form natural deduction, we can prove as follows: Similarly, we can prove: Then, we have the required logical equivalence:


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If you are not allowed to use truth-tables to prove equivalence (even though that's a perfectly acceptable method), but you also don't want to rely on laws like DeMorgan, you could do something like this: Since any conditional $P \to Q$ is False if and only if $P$ is True and $Q$ is False, we have: $P \to (Q \to R)$ is False iff $P$ is True and $Q \to R$ is ...


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(a), (b), and (d) can be proven equivalent by using (five times in total) the fact that $A\implies B$ is equivalent to $(\lnot A)\lor B$, together with de Morgan's law for negating an "and" statement and the associativity and commutativity of "or" statements. Similarly, (c) and (e) can be proven equivalent by using that $A\implies B$ is ...


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The proof you've given is completely choice-free: it just boils down to existential elimination. This can be checked by writing out the fully-formal proof corresponding to that rigorous natural-language argument, and if you want you can even do this in such a way that a computer can verify the final result (although this takes a bit more work). It's worth ...


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Having the answer, I think I can figure out what the question was intended to be. An application of the right-$\land$ rule needs two premises, and those two premises must have the same set $\Gamma$ of formulas to the left of $\vdash$. The two green sequents have the same formulas to the left of $\vdash$, and no two others on your list of five options do that....


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"Firstly, the multiple choices are all of the form Ω⊢F∧G. " By that sort of reasoning, also, all of the multiple choices are of the form Ω⊢F and Ω⊢G. Note that every well-formed formula in formal logic has the form of a variable. But, they are not all of the same form Ω⊢F∧G where Ω indicates a defined list, and they don't all have the same Ω. &...


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The $2$ answers above are indeed excellent, and should be sufficient for your problem. Below I offer a different way of proving this result: $$\begin{align}\sum_{i=1}^{n-1}\frac{1}{i(i+1)}&=\sum_{i=1}^{n-1}\frac{1}{i}-\frac{1}{i+1}\\ &=\frac{1}{1}-\frac{1}{2}\\ &+\frac{1}{2}-\frac{1}{3}\\ &+\frac{1}{3}-\frac{1}{4}\\ &~~\vdots\\ &+\...


1

When $n = 1$, the sum on the left side is empty, because the upper index is $n - 1 = 0$, that is there are no terms to add. This gives $0 = 0$ as desired. Suppose that $$S_n = \sum_{i=1}^{n-1} \frac{1}{i(i+1)},$$ then we have to prove $$S_{n+1} = \frac{n}{ (n+1)}.$$ So if the induction hypothesis $$S_n = \frac{n - 1}{n}$$ is true, we have: $$S_{n+1} =...


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Your understanding of the base case is incorrect. When $n = 1$, the sum on the left-hand side is empty, because the upper index is $n - 1 = 0$, meaning there are no terms to add. This gives $0 = 0$ as desired. The induction step is straightforward. If we let $$S_n = \sum_{i=1}^{n-1} \frac{1}{i(i+1)},$$ then $$S_{n+1} = S_n + \frac{1}{n(n+1)}.$$ So if the ...


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