How can ($A$ and $B$ $\implies$ $C$) and ($C$ and $B$ $\implies$ not $A$) together imply (not $A\iff B$)? I encountered this two statements when I tried to understand the proof of Kuratowski Theorem.


*

*Any minimal nonplanar graph and it has no Kuratowski subgraphs, then it must be at least 3 connected.

*An at least 3 connected graph and it has no Kuratowski subgraphs, then it must be planar.
My question is how can these two statements imply the Kuratowski Theorem?
If we let:
$A=$ minimal nonplanar graph,
$B=$ no Kuratowski subgraphs,
$C=$ at least 3 connected,
Then how can we understand:
$A\wedge B\implies C$
$C\wedge B\implies A'$, where $A'$ is not $A$.
implies
$A'\iff B$
I am not asking about the details about the proof, but only the logic behind how can those two statements imply the third (from the language of logic).
Thanks.
 A: Suppose A is false. Not A implies B. Not A is true. So B holds.
Now suppose B is true. Case 1: C is true. Then C and B are true so not A is true. Case 2: C is false. Then A and B can't be true as A and B imply C. Since B is true, A must be false. In either case, not A is true.
A: Kuratowski's theorem states that:

a finite graph is planar if and only if it does not contain a subgraph that is a subdivision of $K_5$ (the complete graph on five vertices) or of $K_{3,3}$ (complete bipartite graph on six vertices, three of which connect to each of the other three, also known as the utility graph). [...] If $G$ is a graph that contains a subgraph $H$ that is a subdivision of $K_5$ or $K_{3,3}$, then $H$ is known as a Kuratowski subgraph of $G$. With this notation, Kuratowski's theorem can be expressed succinctly: 

a graph is planar if and only if it does not have a Kuratowski subgraph.


If so, we have:
1) "if not planar and no-subgraph, then 3-connected".
2) "if 3-connected and no-subgraph, then planar".
Thus, the "logical form" of 1) is:

$(\lnot PL \land NS) \to 3C$

while for 2) we have:

$(3C \land NS) \to PL$.

While Kuratowski's theorem amounts to:

$PL \leftrightarrow NS$,

where $NS$ stands for: "the graph does not have a Kuratowski subgraph".

We can prove that 1) and 2) imply $NS \to PL$:
1) $(\lnot PL \land NS) \to 3C$ --- premise
2) $(3C \land NS) \to PL$ --- premise
3) $\lnot PL$ --- assumed [a]
4) $NS$ --- assumed [b]
5) $3C$ --- from 3) and 4) by $\land$-intro and $\to$-elim with 1)
6) $PL$ --- from 4) and 5) by $\land$-intro and $\to$-elim with 2)
7) contradiction --- from 3) and 6)
8) $PL$ --- from 3) and 7) by Double Negation, discharging [a]

9) $NS \to PL$ --- from 4) and 8) by $\to$-intro, discharging [b].


But the other "direction": $PL \to NS$ is not implied by 1) and 2).
Consider the case when $PL$ is True and $NS$ is False; we have that 1) is true [$(F \land F) \to ?$ is True] and also 2) is True [a conditional with true cosequent is always True], while $PL \to NS$ is False [$T \to F$ is False].
A: Suppose you have
$$A\text{ and } B\implies C\quad (1)$$
and 
$$C\text{ and } B\implies \text{not } A.\quad (2)$$
Suppose you have $A$, then with the contraposition of $(2)$, you have also not $B$ or not $C$.
If you have not $C$, with the contraposition of $(1)$ you get not $A$ or not $B$. 
Or not $A$ is absurd, so you've not $B$.
If you have instead not $B$, then you've... not $B$.
So in each case you have $$A\implies \text{not }B,$$
which gives you by contraposition $B\implies \text{not } A$.
I don't think the converse is right.
A: Kuratowski's Theorem is an example of TONCAS (The Obvious Necessary Condition is Also Sufficient). That is, the "obvious" necessary condition for a graph to be planar is that it has no Kuratowski subgraph. This is because planarity is a hereditary property (all subgraphs of a planar graph are still planar, so planar graphs can't have nonplanar subgraphs such as $K_{3,3}$ or $K_5$). Hence, it is easy to see that:
$$
\neg A \implies B
$$
The surprising part of Kuratowski's Theorem is the converse, which is what the proof is all about:
$$
B \overset{?}{\implies} \neg A
$$
To this end, suppose that $B$ holds true. We want to show that $\neg A$ holds true. Suppose, towards a contradiction, that $A$ holds true. Then since $A \land B \implies C$, we know that $C$ holds true. But then since $C \land B \implies \neg A$, we know that $\neg A$ holds true, a contradiction. Thus, we conclude that $\neg A$ holds true, as desired.
