One example from Euclidean geometry in planes. Consider the following three axioms:

1. There exist three non-collinear and pairwise different points in the plane.
2. Every 2 distinct points can be connected by a unique line.
3. For every line $g$ in the plane and every point $A$ not on $g$ there exists a unique line $h$ through $A$ (i.e. $A \in h$) with $g \cap h =\emptyset$.

If you assume these axioms to be true, then you can easily prove that the plane consists of at least 4 points in the plane:

Suppose you have a plane. Axiom 1) yields that there are at least 3 non-collinear pairwise different points $A$, $B$ and $C$. According to axiom 2) there is a unique line $g$ through $A$ and $B$. Yet, $C \not \in g$ since $A$, $B$ and $C$ are not collinear due to 1). Now axiom 3) tells you that there is a line $h$ through $C$ with $g \cap h =\emptyset$. According to 2) $h$ is a unique line through (at least) 2 points with one being $C$ by construction. Hence, there has to be a fourth point $D$ in the plane with $D \in h$.

I hope this example helps a bit. I tried to keep it simple and did not use any definitions for a line for example. In a more scientific context this should, however, be done. Since you asked for a toy example the above example should be sufficient.

One example from Euclidean geometry in planes. Consider the following three axioms:

1. There exist three non-collinear and pairwise different points in the plane.

2. Every 2 distinct points can be connected by a unique line.

3. For every line $g$ in the plane and every point $A$ not on $g$ there exists a unique line $h$ through $A$ (i.e. $A \in h$) with $g \cap h =\emptyset$.

If you assume these axioms to be true, then you can easily prove that the plane consists of at least 4 points in the plane:

Suppose you have a plane. Axiom 1) yields that there are at least 3 non-collinear pairwise different points $A$, $B$ and $C$. According to axiom 2) there is a unique line $g$ through $A$ and $B$. Yet, $C \not \in g$ since $A$, $B$ and $C$ are not collinear due to 1). Now axiom 3) tells you that there is a line $h$ through $C$ with $g \cap h =\emptyset$. According to 2) $h$ is a unique line through (at least) 2 points with one being $C$ by construction. Hence, there has to be a fourth point $D$ in the plane with $D \in h$.

I hope this example helps a bit. I tried to keep it simple and did not use any definitions for a line for example. In a more scientific context this should, however, be done. Since you asked for a toy example the above example should be sufficient.

One example from Euclidean geometry in planes. Consider the following three axioms:

1. There exist three non-collinear and pairwise different points in the plane.
2. Every 2 distinct points can be connected by a unique line.
3. For every line $g$ in the plane and every point $A$ not on $g$ there exists a unique line $h$ through $A$ (i.e. $A \in h$) with $g \cap h =\emptyset$.

If you assume these axioms to be true, then you can easily prove that the plane consists of at least 4 points in the plane:

Suppose you have a plane. Axiom 1) yields that there are at least 3 non-collinear pairwise different points $A$, $B$ and $C$. According to axiom 2) there is a unique line $g$ through $A$ and $B$. Yet, $C \not \in g$ since $A$, $B$ and $C$ are not collinear due to 1). Now axiom 3) tells you that there is a line $h$ through $C$ with $g \cap h =\emptyset$. According to 2) $h$ is a unique line through (at least) 2 points with one being $C$ by construction. Hence, there has to be a fourth point $D$ in the plane with $D \in h$.

I hope this example helps a bit. I tried to keep it simple and did not use any definitions for a line for example. In a more scientific context this should, however, be done. Since you asked for a toy example

One example from Euclidean geometry in planes. Consider the following three axioms:

1. There exist three non-collinear and pairwise different points in the plane.
2. Every 2 distinct points can be connected by a unique line.
3. For every line $g$ in the plane and every point $A$ not on $g$ there exists a unique line $h$ through $A$ (i.e. $A \in h$) with $g \cap h =\emptyset$.

If you assume these axioms to be true, then you can easily prove that the plane consists of at least 4 points in the plane:

Suppose you have a plane. Axiom 1) yields that there are at least 3 non-collinear pairwise different points $A$, $B$ and $C$. According to axiom 2) there is a unique line $g$ through $A$ and $B$. Yet, $C \not \in g$ since $A$, $B$ and $C$ are not collinear due to 1). Now axiom 3) tells you that there is a line $h$ through $C$ with $g \cap h =\emptyset$. According to 2) $h$ is a unique line through (at least) 2 points with one being $C$ by construction. Hence, there has to be a fourth point $D$ in the plane with $D \in h$.

I hope this example helps a bit. I tried to keep it simple and did not use any definitions for a line for example. In a more scientific context this should, however, be done. Since you asked for a toy example the above example should be sufficient.