How does conditional independence cause p(y=1,x) = p(y=1,t=1,x)?

Question

I am working through problem set 1 for Stanford's CS 229 course, and I'm stuck on what appears to be a simple question on conditional probability.

Here's the question

If you don't feel like clicking, we're trying to find a s.t. p(t=1|x) = p(y=1|x)/a, where y and x are conditionally independent given t

Using the basic definition of conditional probability, I've been able to do the following: note: I'll be omitting "(i)" superscripts

p(t=1|x) = p(t=1,x)/p(x)

Using p(y=1|t=1,x) = p(y=1,t=1,x)/p(t=1,x) 
=== p(t=1,x) = p(y=1,t=1,x)/p(y=1|t=1,x):

p(t=1|x) = p(y=1,t=1,x)/(p(y=1|t=1)*p(x)) **[equation 1]**

At this point, I'm not seeing a way to isolate p(y=1|x), so I check what I believe to be the answer key. I should probably mention that I am simply following this course via lectures on youtube and I got the problem sets and answer key from somebody's github repo. So far, all answers in the key have matched my own.

Here is the answer according to the key

My issue is with the very first line of this solution

The first line states that p(y=1|x) = p(y=1 | t=1, x) * p(t=1 | x)
If you expand each side using the definition of conditional probability, you'll see that this is basically saying:
p(y=1,x) = p(y=1,t=1,x)

I cannot find any way to verify this

If we assume that this is true and substitute this into equation 1, we get:

p(t=1|x) = p(y=1,x)/(p(y=1|t=1)*p(x))
= p(y=1|x)/p(y=1|t=1)
This gives us alpha = p(y=1|t=1), which is what they got in the answer key (at least according to their second line. The value they define as alpha seems to have come out of thin air.

In case it's not clear, I am extremely confused on how they are getting p(y=1,x) = p(y=1,t=1,x)

In case I'm not considering some critical info, here is the full problem

Answer 1

The first line states that $p(y=1|x) = p(y=1 | t=1, x) * p(t=1 | x)$

If you expand each side using the definition of conditional probability, you'll see that this is basically saying: $p(y=1,x) = p(y=1,t=1,x)$

Yes.

Notice: All labelled examples are positive. $p(t^{(\imath)}=1\mid y^{(\imath)}=1)=1$ .

When given a labelled sample, then it is certain to also be positive. ( For any outcome where $y^{(\imath)}=1$ holds, then too does $t^{(\imath)}=1$. )

Thus the probabilities are equal because they are for the same event. $p(y^{(\imath)}=1, x^{(\imath)})=p(y^{(\imath)}=1,t^{(\imath)}=1,x^{(\imath)})$