I will use the following definitions in the text below:
For each $a,b\in\mathbb{Z}$ let's denote the set of all integers greater than or equal to $a$ and less than or equal to $b$ by $[a,b]_\mathbb{Z}$:
$[a,b]_\mathbb{Z}:=\{z\in\mathbb{Z}|a\leq z \leq b\} = \bigcup_{z=a}^b\{z\}$
Let's denote the set of all natural numbers with $0$ by $\mathbb{N}_0$ :
$\mathbb{N}_0 := \mathbb{N}\cup\{0\}$
And let's denote the set of all functions from $[0,n]_\mathbb{Z}$ to $\mathbb{R}$ by $\mathbb{R}^{[0,n]_\mathbb{Z}}$:
$\mathbb{R}^{[0,n]_\mathbb{Z}} := \{f|f:[0,n]_\mathbb{Z}\to\mathbb{R}\}$
We have to prove the following proposition:
If $n\in\mathbb{N}$ is some natural number, And if $f,g:[0,n]_\mathbb{Z}\to\mathbb{R}$ are two strictly increasing functions with the same image, Then it must be the case that $f=g$
Translating it into logic notation we get that we have to prove:
$\forall n\in\mathbb{N}_0,\forall f,g\in\mathbb{R}^{[0,n]_\mathbb{Z}}, \Big(\forall i\in [0,n-1]_\mathbb{Z}, f(i)<f(i+1)\Big)\land\Big(\forall i\in [0,n-1]_\mathbb{Z},g(i)<g(i+1)\Big)\land f([0,n]_\mathbb{Z})=g([0,n]_\mathbb{Z})\longrightarrow \forall i\in [0,n]_\mathbb{Z},f(i) = g(i)$
We will prove the proposition by induction on $n$:
BASE STEP:
For $n = 0\in\mathbb{N_0}$ we have to prove that:
$\forall f,g\in\mathbb{R}^{\{0\}}, \Big(\forall i\in \emptyset, f(i)<f(i+1)\Big)\land\Big(\forall i\in \emptyset,g(i)<g(i+1)\Big)\land f(\{0\})=g(\{0\})\longrightarrow \forall i\in \{0\},f(i) = g(i)$
(Note: In this case $[0,n]_\mathbb{Z} = [0,0]_\mathbb{Z} = \{0\}$ and $[0,n-1]_\mathbb{Z} = [0,-1]_\mathbb{Z} = \emptyset$)
Let $f,g\in\mathbb{R}^{\{0\}}$, i.e. $f,g:\{0\}\to\mathbb{R}$, be some functions that satisfy:
(1) $\forall i\in \emptyset, f(i)<f(i+1)$
(2) $\forall i\in \emptyset,g(i)<g(i+1)$
(3) $f(\{0\})=g(\{0\})$,
(Side note: (1) and (2) are true by default but we do not need to use any one of them for the proof in this particular case)
We will prove that $\forall i\in \{0\},f(i) = g(i)$:
Since $f(\{0\}) = \{f(0)\}$ and since $g(\{0\})=\{g(0)\}$ we get by (3) that $\{f(0)\}=\{g(0)\}$ and so it must be the case that $f(0)=g(0)$ and we can conclude that $\forall i\in\{0\}, f(i)=g(i)$ as was to be shown.
INDUCTION HYPOTHESIS
Suppose that for some $n = k\in\mathbb{N}_0$ we have
$\forall f,g\in\mathbb{R}^{[0,k]_\mathbb{Z}}, \Big(\forall i\in [0,k-1]_\mathbb{Z}, f(i)<f(i+1)\Big)\land\Big(\forall i\in [0,k-1]_\mathbb{Z},g(i)<g(i+1)\Big)\land f([0,k]_\mathbb{Z})=g([0,k]_\mathbb{Z})\longrightarrow \forall i\in [0,k]_\mathbb{Z},f(i) = g(i)$
INDUCTIVE STEP
We must prove that for $n = k+1$ we have:
$\forall f,g\in\mathbb{R}^{[0,k+1]_\mathbb{Z}}, \Big(\forall i\in [0,k]_\mathbb{Z}, f(i)<f(i+1)\Big)\land\Big(\forall i\in [0,k]_\mathbb{Z},g(i)<g(i+1)\Big)\land f([0,k+1]_\mathbb{Z})=g([0,k+1]_\mathbb{Z})\longrightarrow \forall i\in [0,k+1]_\mathbb{Z},f(i) = g(i)$
Let $f,g\in\mathbb{R}^{[0,k+1]_\mathbb{Z}}$, i.e. $f,g:[0,k+1]_\mathbb{Z}\to\mathbb{R}$, be some functions that satisfy:
(1) $\forall i\in [0,k]_\mathbb{Z}, f(i)<f(i+1)$
(2) $\forall i\in [0,k]_\mathbb{Z},g(i)<g(i+1)$
(3) $f([0,k+1]_\mathbb{Z})=g([0,k+1]_\mathbb{Z}) = A$,
We will prove that $\forall i\in [0,k+1]_\mathbb{Z},f(i) = g(i)$:
By (1) we get $\forall i\in [0,k]_\mathbb{Z}, f(i)<f(k+1)$ and thus $\forall i\in [0,k+1]_\mathbb{Z}, f(i)\leq f(k+1)$, Also since $f(k+1)\in f([0,k+1]_\mathbb{Z})$, We can conclude that $f(k+1)=\max f([0,k+1]_\mathbb{Z})= \max (A)$.
Similarly by (2) we get $\forall i\in [0,k]_\mathbb{Z}, g(i)<g(k+1)$ and thus $\forall i\in [0,k+1]_\mathbb{Z}, g(i)\leq g(k+1)$, Also since $g(k+1)\in g([0,k+1]_\mathbb{Z})$, We can conclude that $g(k+1)=\max g([0,k+1]_\mathbb{Z}) = \max(A)$.
And we've got that:
(4) $f(k+1) = g(k+1)$,
Now let's define two functions $\tilde{f},\tilde{g}\in\mathbb{R}^{[0,k]_\mathbb{Z}}$, i.e. $\tilde{f},\tilde{g}:[0,k]_\mathbb{Z}\to\mathbb{R}$, As follows:
$\forall i\in [0,k]_\mathbb{Z}, \tilde{f}(i) := f(i)$
And $\forall i\in [0,k]_\mathbb{Z}, \tilde{g}(i) := g(i)$
It is clear from (1) and (2) that $\tilde{f}$ and $\tilde{g}$ satisfy:
(5) $\forall i\in [0,k-1]_\mathbb{Z}, \tilde{f}(i)<\tilde{f}(i+1)$
(6) $\forall i\in [0,k-1]_\mathbb{Z}, \tilde{g}(i)<\tilde{g}(i+1)$
Now we'll show that:
(7) $\tilde{f}([0,k]_\mathbb{Z}) = \tilde{g}([0,k]_\mathbb{Z})$
As follows:
$\tilde{f}([0,k]_\mathbb{Z}) = \tilde{f}(\bigcup_{i=0}^k\{i\})=\bigcup_{i=0}^k\tilde{f}(\{i\})=\bigcup_{i=0}^k\{\tilde{f}(i)\}=\bigcup_{i=0}^k\{f(i)\}=\bigcup_{i=0}^{k+1}\{f(i)\}\backslash\{f(k+1)\}=\bigcup_{i=0}^{k+1}f(\{i\})\backslash\{f(k+1)\}=f(\bigcup_{i=0}^{k+1}\{i\})\backslash\{f(k+1)\}=f([0,k+1]_\mathbb{Z})\backslash\{f(k+1)\}$
And now by (3) and (4) we get: $f([0,k+1]_\mathbb{Z})\backslash\{f(k+1)\}=g([0,k+1]_\mathbb{Z})\backslash\{g(k+1)\}=g(\bigcup_{i=0}^{k+1}\{i\})\backslash\{g(k+1)\}=\bigcup_{i=0}^{k+1}g(\{i\})\backslash\{g(k+1)\}=\bigcup_{i=0}^{k+1}\{g(i)\}\backslash\{g(k+1)\}=\bigcup_{i=0}^{k}\{g(i)\}=\bigcup_{i=0}^{k}\{\tilde{g}(i)\}=\bigcup_{i=0}^{k}\tilde{g}(\{i\})=\tilde{g}(\bigcup_{i=0}^{k}\{i\})=\tilde{g}([0,k]_\mathbb{Z})$
Thus by (5), (6) and (7) we can conclude by using the INDUCTION HYPOTHESIS that $\forall i\in [0,k]_\mathbb{Z},\tilde{f}(i) = \tilde{g}(i)$ , But by definition of $\tilde{f}$ and $\tilde{g}$ it implies that $\forall i\in [0,k]_\mathbb{Z},f(i) = g(i)$, And by (4) we get $\forall i\in [0,k+1]_\mathbb{Z},f(i) = g(i)$ as was to be shown.
Q.E.D.
I've proved the proposition and this was the proof I've came up with, Do you think there is a shorter way to prove the proposition? Is this the standard way to prove the proposition? Thanks you...
Your proof is so long and so notation packed that I can't read it to see if it's correct. It probably is.
You ask for a shorter proof. I would argue this way. If $f \ne g$ then there is a smallest $k$ such that $f(k) \ne g(k)$. We can assume $f(k) < g(k)$. Since $g$ is strictly increasing, $f(k)$ can never be in its image. That contradicts the hypothesis so $f$ must be the same as $g$.
(There is really an induction in disguise here, since I've used the fact that any nonempty set of positive integers has a least element.)