English to predicate logic Could someone explain to me why the following"predicate" formulas are the corresponding "english" statement.
Before the actual question. If you could answer these two side question, that would be really helpful. They are referred to as side-question below.
Please try to keep answers moderately detailed, but not too convoluted .As I am not a pro-at -logic.
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Side-question 1
“All men are happy”
Correct solution, below:
∀x(M(x)→ H(x))
Incorrect
∀x(M(x) ∧ H(x))
Why is this incorrect ?
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Side-question 2
“Some men are happy”
Correct solution, below:
∃x(M(x) ∧ H(x))
Incorrect
∃x(M(x)→ H(x))
Why is this incorrect ?
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Actual questions below:

*

*All predicate statements are correct for there respective english statements.

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English statement (1)
" A goal without a plan is just a dream "
Predicate statement (1)
∀x(G(x) ∧ ¬ P(x) → D(x))
Why is the above predicate statement correct for its correspond english statement ?
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English statement (2)
" Rainy days and Mondays always get me down "
Predicate statement (2)
∀x(R(x) v M(x) → G(x))
Why is the above predicate statement correct for its correspond english statement ?
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English statement (3)
“If every circuit works properly and all laptops are functioning then every test-run will terminate.”
Dictionary:
C(x): x is a circuit
W(x): x works properly
L(x) : x is an interface
F(x): x is functioning
R(x): x is a test-run
T(x) : x will terminate
Predicate statement (3)
∀x(C(x) → W(x)) ∧ ∀x(L(x) → F(x)) → ∀x(R(x) → T(x)).
Why isn’t (3) the following correct?
∀x∀y∀z( (C(x) ∧ W(x)) ∧ (I(y) ∧ F(y)) → (R(z) → T(z)) )
Why is the above predicate statement correct for its correspond english statement ?
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 A: It’s understandable you’d find some of these a tad difficult; translating sentences from English to Predicate can sometimes be very challenging, as you need to try and preserve the semantics of each sentence as much as possible. 
Let’s have a look at your first side problem, which has the form:
“All men are happy”.
When it comes to symbolising this sentence, we need to create what’s called an interpretation, which is just the kind of semantics we attach to the symbols (you’ll see what I mean as I go along). In this case, it matters what kind of symbols we use. So, an interpretation of the sentence above looks like this:
Domain:People 
Mx:x is a man 
Hx:x is happy 
In order to come up with a symbolisation for this sentence, one useful way is to gradually introduce the logic into the English sentences, like this:
∀x(if x is a man then x is happy)
∀x(x is a man→ x is happy)
∀x(Mx→Hx)   
Now, the symbolisation of the sentence is:
∀x(Mx→Hx) 
Breaking a sentence up into its semantic components is a good way to understand what a sentence is actually saying. I think this might help explain why your symbolisation of this sentence as ∀x(Mx∧Hx) is incorrect. If we apply this process backwards, we get:
∀x (x is a man∧  x is happy)
∀x(x is a man and x is happy)
Everyone is a happy man                    
It’s clear that “Everyone is a happy man” is not semantically equivalent to “All men are happy”.  In the former case, we need to specify the fact that the property of “happiness” only applies when the person is a man, not when they could be anybody. In the latter case, we are essentially saying that everything within the scope of the domain (namely people) possesses the properties of “happy” and “man”, which is clearly not what this sentence means. But we need to make an important observation. In almost all cases (and by that I mean cases where the English sentence actually makes sense), the main operator inside a universal quantifier is a conditional statement, whereas the main operator inside an existential quantifier is generally an and statement. The reason for this is that in the English language, we tend to refer to Some x is y to mean ‘there is at least one x and it has the property y’, whilst we refer to All x is y to mean ‘everything that is in x must also be in y’. The word ‘and’ in the former sentence pretty clearly denotes the logical hat symbol, but the phrase ‘must also be’ is actually equivalent, at least in meaning, to the conditional. 
I will assume that you understand my reasoning on your first side example and proceed to do the same from now on, with the same logical analysis. Now let’s have a look at your second side example, which faces the same problem I’ve just indicated above (namely a conditional inside an existential):
“Some men are happy”.
The construction of the symbolisation looks like this:
∃x(x is a man and x is happy)
∃x(x is a man∧ x is happy)
∃x(Mx∧Hx)                                      
Now, the symbolisation of the sentence is:
∃x(Mx∧Hx) 
But a deconstruction of  ∃x(Mx→Hx) looks like this:
∃x (x is a man→  x is happy)
∃x(if x is a man then x is happy)
There is a person who is happy if they are a man.
Clearly, “There is a person who is happy if they are a man” is not the same as our original sentence. In the former case, it states that some members of the set of men are happy; this sentence allows for the fact that some men may not be happy. In the latter, someone is a man
only when they are among the set of happy people; this sentence does not allow for the fact that some men may not be happy. 
Now let’s tackle your so-called Actual Questions. 
*QUICK NOTE: “All predicate statements are correct for their respective English statements”. You’ll actually see that this isn’t quite true, and that there are small errors in some of the symbolisations.
1) “A goal without a plan is just a dream”
Our interpretation is:
Domain:People 
Gx:x is a goal 
Px:x is a plan 
Dx:x is a dream 
Our symbolisation is:
∀x(if x is a goal and does not have a plan then x is just a dream)
∀x(if x is a goal ∧ does not have a plan then x is just a dream)
∀x(( x is a goal ∧ does not have a plan)→ x is just a dream)
∀x(( Gx ∧ ¬Px)→ ~Dx)                      
A couple of notes. The original English sentence is incredibly ambiguous. It is unclear whether ‘A goal” refers to a single goal, or if it implicitly refers to ‘Any goal’.  For simplicity’s sake, I have assumed the latter, as an assumption of the former would produce a different formula. The second point I want to make is the use of brackets; it is also incorrect. The only instances when you can omit brackets are as follows:
*Successive application of ∨, i.e. ∨…∨
*Successive application of ∧, i.e. ∧…∧
*Any outer brackets
Whilst these conditions do not apply to Predicate in the strictest sense, the reason why it is okay to do so in a more relaxed sense is because none of these rules affect the truth-functional properties of a statement. Any other bracket omission does, and unfortunately, your ‘correct’ solution proves somewhat ambiguous. Does it mean 
∀x(( Rx ∨ Mx)→ ~Gx) or ∀x(Rx ∨( Mx→ ~Gx))? So, the correct solution proves to be ∀x(( Rx ∧ Mx)→ ~Gx), and from now on I will take the problem of brackets as a given.     
2) “Rainy Days and Mondays always get me down”
Our interpretation is:
Domain:Days and Emotions
Rx:x is a rainy day 
Mx:x is monday(s) 
Gx:x is gets me down 
Our symbolisation is:
∀x(if x is a rainy day and x is monday then x gets me down)
∀x(if ( x is a rainy day ∧ x is monday)  then x gets me down)
∀x(( x is a rainy day ∧ x is monday)→ x gets me down)
∀x(( Rx ∧ Mx)→ Gx)       
Unfortunately, your supposedly ‘correct’ statement is in fact incorrect again, but for different reasons this time. The main issue here is the use of ∨ instead of ∧. The former refers to English statements that feature something like “Either…or…”; but this phrase is clearly absent from this sentence. The use of ‘and’ in the sentence is a very clear indication that we should apply ∧, as this is what this particular symbol denotes (at least in English). 
3) “If every circuit works properly and all laptops are functioning then every test run will terminate”
Our interpretation is:
Domain:Technology 
Cx:x is a circuit 
Wx:x works properly 
Lx:x is a laptop 
Fx:x is functioning 
Rx:x is a test run 
Tx:x is will terminate 
Our symbolisation is:
*if for every x, if  x is a circuit then x works properly and every if y is a laptop then it is also functioning then if every test runs then it will terminate
    (for every x,if  x is a circuit then x works properly and for every y,if y is a laptop then it is 
also functioning)→(if every test runs then it will terminate)
*(for every x(x is a circuit→x works properly) and for every y(y is a laptop→it is also functioning)→(if every test runs then it will terminate)
*(∀x(x is a circuit→x works properly)∧∀y(y is a laptop→it is also functioning)→∀z(z is a test runs→it will terminate)
*(∀x(Cx→Wx)∧∀y(Ly→Fy)→∀z(Rz→Tz)      
∀x∀y∀z(((Cx ∧ Wx ∧ Ly ∧Fy)→(Rz→Tz))) has one main problem; the main operator of the English sentence is very evidently the conditional arrow and not the universal quantifier, which is nested within the conditional. So, barring bracket errors, (∀x(Cx→Wx)∧∀y(Ly→Fy))→∀z(Rz→Tz) is the correct symbolisation for this sentence. I should just say that unlike the symbolisation that you’ve given, I have assumed that the variables x, y, z are distinct. This is because nowhere in the sentence does it explicitly specify that x, y, z are not distinct, so I think it would be a disservice to the semantics to assume this (other than that, and the brackets of course, my formula is the same as yours). 
I sincerely hope some of this helped, even if just a little. Feel free to ask if not, especially since I used more notation and introduced some new ideas. 
Final note: If you’re looking for more examples of English to Predicate, I’d recommend looking at forallx; it’s an open-source textbook, which is handy in understanding Propositional and Predicate logic in a less technical format. 
A: Quantified predicate logic (FOL) is certainly much more technical and hard to understand compared with propositional calculus where you only have familiar truth-value connectives such as usual not, and, or, if/then. There's a critical terminology called extension. All propositional connectives are extensional such as negation: ¬A holds $\iff$ A does not; so whether ¬A holds only depends on whether A holds or doesn’t, so you can use truth table in a finite manner. But quantifiers are not extensional, for the truth value of ∃xP(x) doesn't depend on the truth value of P(x) since since P(x) contains a free variable thus it is neither true nor false. So you have to go into the semantic meaning of the quantifier, and as explained by Angus St. before you'll soon understand all your main questions.
Final note on your last question where you have a prenex normal form which you're unsure about its natural language translation correctness. Every classical first-order formula is logically equivalent to some formula in prenex normal form (not for intuitionistic logic though). There are several conversion rules that can be recursively applied to convert a formula to prenex form such as quantified DeMorgan's rule, null quantification rules, etc. Let's start your statement (3) in a Fitch like natural deduction style:
|∀x(C(x) → W(x)) ∧ ∀x(L(x) → F(x)) → ∀x(R(x) → T(x))
|∀x(C(x) → W(x)) ∧ ∀y(L(y) → F(y)) → ∀z(R(z) → T(z))
|∀x[(C(x) → W(x)) ∧ ∀y(L(y) → F(y)] → ∀z(R(z) → T(z))        ...Null Quantification on ∀x
|∀x∀y[(C(x) → W(x)) ∧ (L(y) → F(y)] → ∀z(R(z) → T(z))        ...Null Quantification on ∀y
|¬∀x∀y[(C(x) → W(x)) ∧ (L(y) → F(y)] ∨ ∀z(R(z) → T(z))       ...→ replacement
|∃x∃y[¬((C(x) → W(x)) ∧ (L(y) → F(y))] ∨ ∀z(R(z) → T(z))     ...Quantified DeMorgan's applied twice consecutively
|∀z∃x∃y[¬(((C(x) → W(x)) ∧ (L(y) → F(y))) ∨ ((R(z) → T(z)]     ...Null Quantification on ∀z
|∀z∃x∃y[((C(x) → W(x)) ∧ (L(y) → F(y)) → ((R(z) → T(z))]      ...→ replacement
So the correct prenex form is only slightly different than yours for the order of the 3 quantifiers and their respective types.
