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How can I prove if a set of first order logic formulas is satisfiable or not?

For example, if I have the following set:

$\{∃x∀y¬R(y,x),∀yR(y,y)\}$

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    In this case, just use the definition of satisfiable: to show that it is satisfiable, show that there is a model with an interpretation of $R$ in which the sentences hold; and to show that it is not satisfiable, show that in no model with an interpretation of $R$, the sentences both hold. – 2017-01-02
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    So basically I just need to find a set $A$ and an interpretation of $R^A$ where the formulas are both true and that's my proof done? – 2017-01-02
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    Yes, *if* the set of formulas *is* satisfiable, then that is a proof. Of course, if they are not, that will not be impossible. – 2017-01-02
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    If I say that $A=\{1,2,3\}$ and $R^A=\{(1,1),(2,2),(3,3)\}$ the formulas are both true, correct? – 2017-01-02
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    No, the first one is not satisfied. To respond to your comment below, $x$ and $y$ can indeed have the same value. – 2017-01-02

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Hint

For unsatisfiability, you can try to find a contradiction.

The first premise implies, for some $a$:

$∀y¬R(y,a)$.

Then, instantiating both the above formula and the second premise with $a$, we get :

$¬R(a,a)$ and $R(a,a)$

a contradiction.

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    If I say that $A = \{1,2,3\}$ and $R^A = \{(1,1),(2,2),(3,3)\}$ the set is satisfiable, correct? – 2017-01-02
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    @JoãoSilva - No: $∃x∀y¬R(y,x)$ is not. – 2017-01-02
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    But if $y=1$ and $x=3$ the sentence is not true, and the same thing happens when $y=2$ and $y=3$. Question: is $∃x∀y¬R(y,x)$ the same as $∀y∃x¬R(y,x)$? – 2017-01-02
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    Wait, can $x$ have the same value of $y$? – 2017-01-02
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    @JoãoSilva - "\forall" means **for all**. – 2017-01-02
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    I don't understand why $∃x∀y¬R(y,x)$ is not satisfiable given the structure above. Can you give me an example where it is not true? – 2017-01-02
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    I understand now. Let's say that my set is $\{∃xR(x),∀x¬R(x)\}$ I can say that it isn't satisfiable because if $A = \mathbb{N}$ and $R^A = \{n \in \mathbb{N} | n is even\}$ then $∀x¬R(x)$ is not true, correct? – 2017-01-02
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    @JoãoSilva - Correct, but... to show that it is *unsatisfiable* you have to prove that there is **no** interpretation where it is true. Thus, it does not suffice to consider one interpretation. In the case of your comment, you have to consider that $∀x¬R(x)$ is equiv to $¬∃xR(x)$, and thus the set with the two formulas is inconsistent, i.e. unsat. – 2017-01-02
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    @JoãoSilva - you are welcome :-) – 2017-01-02