MHB Proving ${\psi}_{n}(x)\le F(n)$ by Induction

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The discussion focuses on proving by induction that ${\psi}_{n}(x) \le F(n)$ for $x \in [a, b]$. The user establishes a base case for $n=1$, confirming that ${\psi}_{1}(x) \le f(x,1) \le F(1)$. They also demonstrate that if ${\psi}_{n}(x) \le f(x,n)$ holds, then it follows that ${\psi}_{n+1}(x) \le F(n+1)$. The user seeks clarification on whether this implies that ${\psi}_{n}(x) \le F(n)$ is valid. Definitions for ${\psi}_n$, $F$, $a$, and $b$ are requested to further the discussion.
sarrah1
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Hello

is my proof be correct ?

I wish to prove by induction that ${\psi}_{n}(x)\le F(n)$ , $x\in[a,b]$ ... (1)

Let there exists a function $f(x,n)$ such that if ${\psi}_{n}(x)\le f(x,n) $ then ${\psi}_{n}(x) \le F(n)$ .

I know that (1) is true for $n=1$ i.e. ${\psi}_{1}(x)\le f(x,1)\le F(1)$ ,

and I was able to prove that

${\psi}_{n+1}(x)\le F(n+1)$ , $x\in[a,b]$

would this implies ${\psi}_{n}(x)\le F(n)$

thanks
 
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sarrah said:
I wish to prove by induction that ${\psi}_{n}(x)\le F(n)$ , $x\in[a,b]$
Please provide the definitions of $\psi_n$, $F$, $a$ and $b$.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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