Not understanding this way of proving undecidability of the termination problem

I am reading some slides on Algorithm to understand why termination is an undecidable problem. The slides say the following:

– Assume termination(P) always terminates and returns true iff P always terminates on all input data;

– The following program yields a contradiction

while termination(P) skip;  

But I do not see what kind of contradiction there. Any idea?

Want help with proving a calculus theory

Let $ f(x)$ have a second derivative on the closed interval $ [-2,2]$ . If $ \left| f(x) \right| \le 1$ and $ \frac{1}{2} (f^{\prime}(0))^2+f(0)^3>\frac{3}{2} $ when $ -2\le x\le2$ , now I need to prove that there must be a point $ x_{0}$ on the interval $ (-2,2)$ such that $ f^{\prime \prime}\left(x_{0}\right)+3\left[f\left(x_{0}\right)\right]=0$ .

(Series[1/2 (f'[x])^2 + f[x]^3, {x, 0, 1}]) // FullSimplify  1/2 (Series[f'[x], {x, 0, 1}] // Normal)^2 + (Series[      f[x], {x, 0, 1}] // Normal)^3 // Expand  (Series[f''[x], {x, 0, 1}] // Normal)^2 + (3*Series[f[x], {x, 0, 1}] //      Normal)^2 // Expand 

The above code does not reveal the nature of the problem and solve it cleverly. What can I do to solve this problem?

Problem with proving that $RP \subseteq NP$ : a non-deterministic TM for a language $L \in RP$

I’m having a small issue with wikipedia’s proof that $ RP \subseteq NP$ :

An alternative characterization of RP that is sometimes easier to use is the set of problems recognizable by nondeterministic Turing machines where the machine accepts if and only if at least some constant fraction of the computation paths, independent of the input size, accept. NP on the other hand, needs only one accepting path, which could constitute an exponentially small fraction of the paths. This characterization makes the fact that RP is a subset of NP obvious.

My problem lies in the following statement: "…accepts if and only if at least some constant fraction of the computation paths, independent of the input size, accept"

My question is: for a language $ L \in NP$ , wouldn’t it only need one computation path for the non-deterministic TM to accept? Since, if $ x \in L$ , $ P[T(x) accepts] >= 1/2$ with $ T$ being the Turing machine that decides $ L$ .

Why is the proof complete after proving only for one induction when we have more than one variable?

So I’m self-learning coq. And I came across the proof for associativity in addition forall (a b c : nat)

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Appearntly when we do induction a. after intros a b c., it creates 2 subgoals

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and afterwards we simply need to show that the two sides in both subgoals are equivalent, and the proof is completed.

So I’m just wondering why we don’t need to do induction b. and induction c. to complete the proof? Why only performing induction on a is able to complete the proof?

Or in other words, how come in the function that returns the proof, we just get b and c for “free”? Constructively don’t we need something like a double induction applied twice?

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Proving injectivity for an algorithm computing a function between sets of different types of partitions [closed]

I am attempting to solve the following problem:

Let $ A$ be the set of partitions of $ n$ with elements $ (a_1, \dots, a_s)$ such that $ a_i > a_{i+1}+a_{i+2}$ for all $ i < s,$ taking $ a_{s+1} = 0.$ Define $ g_n = F_{n+2}-1$ and let $ B$ be the set of partitions of $ n$ as $ b_1 \ge \dots \ge b_s$ such that $ b_i \in \{g_n\}$ for all $ i,$ and if $ b_1 = g_k$ for some $ k,$ then $ g_1, \dots, g_k$ all appear as some $ b_i.$ Prove $ |A|=|B|.$

Attempt: Let $ e_i$ be the vector with $ 1$ at position $ i$ and $ 0$ elsewhere. If $ b_1 = g_k,$ let $ c=(c_k, \dots, c_1)$ count how many times $ g_i$ appears. We calculate $ f: B \to A$ as follows:

Let $ c=(c_k,\dots,c_1), a=(0,\dots,0).$ While $ c \ne 0,$ let $ d_1 > \dots > d_k$ be the indices such that $ c_{d_i} \ne 0.$ Replace $ c, a$ with $ c-(e_{d_1}+\dots+e_{d_k}), a+(g_{d_1} e_1 + \dots + g_{d_k} e_k)$ respectively. After the while loop ends, let $ f(b)=a.$

Let $ \sum a, \sum b, \sum c$ be the sum of the components of $ a, b, c$ respectively. Since $ \sum c$ decreases after every loop, the algorithm terminates and $ f(b)$ is well-defined. Since $ c_k g_k + \dots + c_1 g_1 + \sum a$ does not change after every iteration, is $ \sum b$ at the start and $ \sum a$ at the end, we have $ \sum f(b) = \sum b = n,$ so $ f(b)$ is also a partition of $ n.$ Now $ a = (g_k, \dots, g_1)$ after the first loop, which satisfies the condition $ g_i > g_{i-1}+g_{i-2}$ since $ g_i = F_{n+2}-1 = (F_{n+1}-1)+(F_n-1)+1 > g_{i-1}+g_{i-2}.$ Furthermore, after every iteration of the loop, the difference $ a_i – (a_{i-1}+a_{i-2})$ changes by $ 0, g_{d_j}-g_{d_{j-1}} > 0,$ or $ g_{d_j}-(g_{d_{j-1}}+g_{d_{j-2}}) > 0,$ so we have $ a_i > a_{i-1} + a_{i-2}$ at the end and hence $ f(b) \in A.$ Thus, $ f: B \to A$ is well-defined.

In order to prove the injectivity of $ f,$ it suffices to prove each loop iteration as a mapping $ (c,a) \to (c’,a’)$ is injective, which would imply the mapping $ (c,0) \to (0,a)$ that the while loop creates is injective. Indeed, if $ f(b_1) = f(b_2) = a$ with $ (c_1, 0), (c_2, 0)$ being sent to $ (0, f(b_1)) = (0,a), (0, f(b_2)) = (0,a)$ respectively, then we have $ (c_1, 0) = (c_2, 0) \Rightarrow c_1 = c_2 \Rightarrow b_1 = b_2.$

Suppose $ d_1 > \dots > d_i, f_1 > \dots > f_j$ are the non-zero indices of $ c_1, c_2$ respectively and $ c_1 – (e_{d_1}+\dots+e_{d_i}) = c_2 – (e_{f_1}+\dots+e_{f_j}), a_1+g_{d_1}e_1 + \dots+ g_{d_i} e_i = a_2 + g_{f_1} e_1 + \dots + g_{f_j} e_j.$ If $ x \ge 2$ is an entry of $ c_1,$ it decreases by $ 1,$ so the corresponding entry in $ c_2$ after $ c_2$ is modified is also $ x-1,$ which means it must’ve been $ (x-1)+1 = x$ before since $ x-1>0.$ Thus, if the values of two positions of $ c_1, c_2$ differ, one is $ 1$ and the other is $ 0.$ However, if $ c_1 = (1,0), a_1 = (3,1), c_2 = (0,1), a_2 = (4,1),$ then $ (a_1, c_1), (a_2, c_2)$ both get sent to $ ((5,1), (0,0)).$ I can rule out this specific example by arguing that one of the pairs is illegal and could not have come from any choice of initial $ c,$ but I have no idea on how to do this in general.

What should I do next in order to show $ f$ is injective? Furthermore, since the problem I’m trying to prove is correct, injectivity would imply $ f$ is secretly a bijection. But I have no clue on how to even start on the surjectivity of $ f,$ so I just constructed a similar algorithm for $ g: A \to B$ in the hopes of proving $ g$ is injective too. If I can show $ f$ is injective I will probably know how to show $ g$ is.

Here is an example of $ f, g$ in practice:

Let $ n = 41, b = (12, 7, 7, 4, 4, 2, 2, 2, 1) \Rightarrow c = (1, 2, 2, 3, 1).$

$ $ ((1, 2, 2, 3, 1), (0,0,0,0,0)) \to ((0, 1, 1, 2, 0), (12, 7, 4, 2, 1)) \to ((0, 0, 0, 1, 0), (19,11,6,2,1)) \to ((21,11,6,2,1),(0,0,0,0,0)),$ $ so $ f(b) = (21,11,6,2,1).$

Let $ a = (21, 11, 6, 2, 1).$

$ $ ((21,11,6,2,1),(0,0,0,0,0)) \to ((9,4,2,0,0), (1,1,1,1,1)) \to ((2,0,0,0,0),(1,2,2,2,1)) \to ((0,0,0,0,0),(1,2,2,3,1)),$ $ so $ g(a) = (12, 7, 7, 4, 4, 2, 2, 2, 1).$

Proving a certain primitive recursive function exists

Assume $ f\colon ω × ω → ω$ is a computable function. How can we prove that there is a primitive recursive function $ g\colon ω × ω → ω$ where the following holds:

$ ∀n [∃s(f(n, s) = 1) ↔ ∃k(g(n, k) = 1)]$

So for every $ n$ , there is an $ s$ such that $ f(n, s) = 1$ if and only if there is a $ k$ such that $ g(n, k) = 1$ .

Been working on this problem for a while now, if anyone could please help?