My premises are:
$\forall x(x\in A \leftrightarrow(x=R \lor \exists y(y\in A \land x=y^{+1})))$
$\forall p\forall r\forall x(\langle p,r\rangle \in x^{+1} \leftrightarrow \exists q(\langle p,q\rangle\in x \land \langle q,r\rangle \in R))$
$\langle a,b\rangle\in R$
$\langle b,c\rangle \in B \land B\in A$
And, I want to prove the following:
- $\exists z(\langle a,c\rangle \in z \land z\in A)$
The symbols I am using are:
$$ \begin{array}{lr} \text{2-place Predicate} & \in \\ \text{1-place Function} & ^{+1} \\ \text{2-place Function} & \langle \rangle \\ \text{Constants} & R,A,B,a,b,c \end{array} $$
Meaning
We can have different meanings of the above premises and conclusion, but there is a primary meaning in my mind.
We can assume that the domain of discourse contains sets and other objects. So, we can think of $a$,$b$ and $c$ as some objects,$\langle a,b\rangle$ as an ordered pair and $R$ to be a set.
Premise (1) describes set $A$ recursively:
- $A$ is a set whose elements are only $R$,$R^{+1}$,$R^{+1+1}$,$R^{+1+1+1}$....
Premise (2),(3) and (4) says:
For any set $x$, $x^{+1}$ is a set and it contains $\langle p,r\rangle$ iff there is a $q$, such that $\langle p,q\rangle$ belongs to $x$ and $\langle q,r\rangle$ belongs to $R$
$R$ contains $\langle a,b\rangle$
$B$ is a set containing $\langle b,c\rangle$. Also, $B$ belongs to $A$.
Now, the conclusion is saying :
- There exists a $z$, such that $\langle a,c\rangle$ belongs to $z$ and $z$ belongs to $A$.
Intuitive Explaination
As $B$ belongs to $A$, $B$ is equal to $R$ or $R^{+1}$ or $R^{+1+1}$ or $R^{+1+1+1}$... and for each value of $B$ we can show that the conclusion holds.
For example, assume $B=R^{+1+1}$
So, $\langle b,c\rangle$ belongs to $R^{+1+1}$ since $\langle b,c\rangle$ belongs to $B$
Since $\langle b,c\rangle$ belongs to $R^{+1+1}$, there exists a $d$ such that $\langle b,d\rangle$ belongs to $R^{+1}$ and $\langle d,c \rangle$ belongs to $R$ (by premise (2))
Since $\langle b,d \rangle$ belongs to $R^{+1}$, there exists a $e$ such that $\langle b,e\rangle$ belongs to $R$ and $\langle e,d\rangle$ belongs to $R$
Now, since $\langle a,b\rangle$ and $\langle b,e\rangle$ belongs to $R$, $\langle a,e\rangle$ belongs to $R^{+1}$
Since $\langle a,e \rangle$ belongs to $R^{+1}$ and $\langle e,d\rangle $ belongs to $R$, $\langle a,d\rangle$ belongs to $R^{+1+1}$
Since $\langle a,d\rangle$ belongs to $R^{+1+1}$ and $\langle d,c\rangle$ belongs to $R$, $\langle a,c\rangle$ belongs to $R^{+1+1+1}$
So, $\langle a,c\rangle$ belongs to $R^{+1+1+1}$ and $R^{+1+1+1}$ belongs to $A$
So, There exists a $z$, such that $\langle a,c\rangle$ belongs to $z$ and $z$ belongs to $A$.
Where I am stuck
In the last step, I reached $C\in A \land B=C^{+1}$. Now, I should reach $C=R \lor \exists y(y\in A \land C=y^{+1}) $, I can prove the conclusion for $C=R$ case but for $\exists y (y\in A \land C=y^{+1} $, I have to introduce another constant $D$. But then again for $D$ I will get the case $\exists y(y\in A \land D=y^{+1})$ and so have to introduce another constant. So, I am stuck in 'infinite steps'.