Equivalence between Minimality Conidition, Well-Founded Property, Descending-Chain-Condition, and Noertherin Induction Let $(M, \preceq)$ denote a partially ordered set $M$ along with a partial order $\preceq$ on it.
Proof of the equivalence between: A: Descending Chain Condition
If ($\mathcal{C}$ is a decresasing chain in M) $\rightarrow$ ($\mathcal{C}$ is finite)
More formally: For any chain  $a_o  \succeq a_1 \succeq ... a_i \succeq ...$, there exists a number n such that $a_n = a_{n+1} =...$
B: Well-Founded-Property: There does not exist an infinite strictly decreasing chain in M
C: Minimality Property: Every non-empty subset in M has a minimal element
condition 1) P(x) is true for all minimal elements x $\in$ M 
condition 2) If P(z) is true $\forall$ z $\in$ M such that z $<$ y, for some y $\in$ M $\rightarrow$ P(y)

D: Noertherian Induction Property: (condition1 $\wedge$ condition2) $\rightarrow$ P(a) $\forall a \in M$
We will first prove $A \rightarrow B$
Proof by contradiction

Assume $A$ $ \wedge \neg B$
Then we are assuming A, as well as the existence of an infinite
  descending chain in M. Let us call this chain $\mathcal{C}$ But by A,
  since $\mathcal{C}$ is a descending chain it should be finite, but by
  our assumption of $\neg B$ we are assuming otherwise. 
Contradiction

proof of $B \rightarrow C$
 Proof by contradiction

Assume $B$ $\wedge \neg C$
Let $\mathcal{C}$ be a non-empty subset of M such that there is no
  minimal element in $\mathcal{C}$. This set exists by assumption of
  $\neg C$.
Since $\mathcal{C}$ is not empty, there exists an $a_0 \in
\mathcal{C}$. 
If there does not exist a second element $a_1 \in \mathcal{C}$, then
  $a_0$ would constitute a minimal element (w.r.t $\mathcal{C}$).  This
  would contradict assumption $C$.
Then $\mathcal{C}$ must contain more than one element. 
Since  $\mathcal{C}$ does not contain a minimal element, there must
  exist $a_1$ in $\mathcal{C}$ such that $a_o  \succ a_1$ Since $a_1$
  cannot be a minimal element there must exist $a_2$ in $\mathcal{C}$
  such that  $a_o  \succ a_1 \succ a_2 $
If there exists an $a_k$ in $\mathcal{C}$ such that $a_0  \succ a_1
 \succ   a_2 \succ ... \succ a_k$, for all $a_i \neq a_k \in
 \mathcal{C}$,then $a_k$ would constitue a minimal element. This
  would contradict assumption $B$.
Therefore, $\mathcal{C}$ does not have a minimal element, yet
  constitues an infinite strictly descending chain. 
This is our final contradiction, as this contradicts assumption $B$.

We want to now prove $ C \rightarrow D$
Proof by contradiction

Assume $C \wedge \neg D $ Then we are assuming $C$ $ \wedge $
  [condition1 $\wedge$ condition2] $\wedge$ $ [\exists  a \in  M$ such
  that $\neg$P(a)]
Let  $P(x)$ be a propositional function that satisfies conditions 1
  and 2
Let $ \mathcal{X} = \{ x \in M \, | \, \neg P(x) \}.$
Our assumption of $\neg D$ assumes that $ \mathcal{X}$ is not empty 
Then by assumption $C$, every non-empty set has a minimal element.
  Thus there exists a minimal element, $x_0$ in $\mathcal{X}$
Since we are assuming condition1, P($x_0$) must hold.
This yields a contradiction as P($x_0$) holds, yet $x_0$ is in 
  $\mathcal{X}$ which is defined to be such that $\neg$ P($x$) holds for
  all elements which are members.
The contradiction is that P($x_0$) and $\neg$ P($x_0$) cannot both
  hold.
Therefore, $\mathcal{X}$ cannot be non-empty, and therefore is empty.
  Thus, our condition of  $\neg D $, that of: [condition1 $\wedge$
  condition2] $\wedge$ $ [\exists  a \in  M$ such that $\neg$P(a)], is
  contradicted

We want to now prove $ D \rightarrow A$
Proof by contradiction

Assume $D \wedge \neg A $ Then we are assuming the Noertherian
  Induction Property as well as the negation of the Descending Chain
  Condition That is, we are assuming the Noertherian Induction Property
  and If ($\mathcal{C}$ is a decresasing chain in M) $\rightarrow$
  ($\mathcal{C}$ is finite)

I am stuck for this last part of the proof.
So is what I have so far ok, and also can anyone help with the remaining piece?
Thanks
A: Here is a hint for $D\implies A$: We are going to prove its contraposition. Assume that there is an infinite decreasing sequence $A=\{a_1,a_2,\cdots\}$ and consider the proposition $x\not\in A$. Every minimal element is not in $A$. Moreover if $y\in A$ then it has a predecessor $z\in A$, and its contrapositive is the condition 2 of your statement of Noetherian induction. However not every element satisfies $x\notin A$.
