Formal Proofs of Logic I need to give Fitch-style formal proofs for the following:
1)  Premises:


*

*∀x∀y∀z((R(x, y) ∧ R(y, z)) → R(x, z))

*∀x∀y(R(x, y) → R(y, x))


To prove: ∀x∀y(R(x, y) → R(x, x))
2) Premises:


*

*∀x(P(x) → Q(a))

*∃x(P(x) ∧ Q(x))


To prove: Q(a)
For question 1 I tried something, but it became one big mess with a lot of subproofs. The amount of different variables is throwing me a bit off. For question 2 I can't even see why the goal sentence follows from the premises.
Appreciating all input.
 A: Without delving into proper Fitchean formalism, you can proceed as follows.
1)
The premises say that $R$ is transitive and $R$ is symmetric. The conclusion says that (for all $x$) if $x$ bears $R$ to anything than it bears $R$ to itself.


*

*Suppose $x, y$ are such that $R(x,y)$

*By 2., $R(x,y) \to R(y,x)$, hence modus ponens applied to this with the previous step yields:

*$R(y,x)$

*By 1., $R(x,y) \wedge R(y,x) \to R(x,x)$, so

*$R(x,x)$


Because $x,y$ were arbitrary, 


*

*$\forall x R(x,x)$


2)
Given 1. and 2.


*

*By 2. there is something $b$ such that $P(b) \wedge Q(b)$, so 

*$P(b)$

*In 1. taking $x = b$, we get $P(b) \to Q(a)$; thus, by modus ponens,

*$Q(a)$

A: 
1)  Premises:
  
  
*
  
*∀x∀y∀z((R(x, y) ∧ R(y, z)) → R(x, z))
  
*∀x∀y(R(x, y) → R(y, x))
  
  
  To prove: ∀x∀y(R(x, y) → R(x, x))
I tried something, but it became one big mess with a lot of subproofs. The amount of different variables is throwing me a bit off. 

Hint: $\forall x\forall y\forall z \; P(x,y,z) \vdash P(a,b,a)$ by universal instantiation.  They don't have to be three arbitrary variables, you can weaken it as needed.
Essentially: apply universal instantiation on the premises to obtain predicates of two arbitrary witnesses, $a,b$, then demonstrate using these results that assuming $R(a,b)$ concludes $R(a,a)$ by modus ponens (twice), and then apply universal generalisation to that result.


2) Premises:
  
  
*
  
*∀x(P(x) → Q(a))
  
*∃x(P(x) ∧ Q(x))
  
  
  To prove: Q(a)

Use existential generalisation then conjunctive elimination on the second premise to obtain the witness $P(c)$, then use universal generalisation on the first premise to show $P(c)\to Q(a)$ and put them together using modus ponens to conclude $Q(a)$.
A: Using hints from the other answers here are Fitch-style proofs in a proof checker to show the details:



Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
