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  1. ~q ⋀ r ⋀ (~s -> t) using only ∨ , ~ Ive got

~(~(q ∨ ~ r) ∨ (s ∨ t)

but this is wrong according to the truth tables

  1. p ∨ (~q ⋀ (r ⩖[exclusive or] s)) using only -> and ~

    This one completely stumped me

If there is justice in this life, then there is no need for a future life. If, on the other hand, there is no justice in our earthly life, then we have no reason to believe that God is righteous. But if we have no reason to believe that God is righteous, then we have no reason to believe that He will provide us with a future life. So either we don't need a future life, or we have no reason to believe that God will provide us with such a life.

If there is justice in this life-p

then there is no need for a future life.-q

If, on the other hand, there is no justice in our earthly life,- ~p

we have no reason to believe that God is righteous -r

we have no reason to believe that He will provide us with a future life - s

we don't need a future life -q

or we have no reason to believe that God will provide us with such a life. -s

I dont know how to connect the sentences here

ive got p->q

    ~p->r

     r->s
    
     q⩖(exclusive or) s
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  • $\begingroup$ You simply misplaced one of the ~s in the first question. $\endgroup$ – player3236 Jan 21 at 17:47
  • $\begingroup$ In number 2, what is that superimposed double AND symbol? $\endgroup$ – fleablood Jan 21 at 17:55
  • $\begingroup$ @fleablood exclusive or $\endgroup$ – wedsdf qwesq Jan 21 at 17:55
  • $\begingroup$ @player3236 can you elaborate , please ? $\endgroup$ – wedsdf qwesq Jan 21 at 17:56
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    $\begingroup$ @CSchofx Sorry . In the country i come from that is the symbol we use . $\endgroup$ – wedsdf qwesq Jan 21 at 17:58
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Question 2. There might be a better way, or shorter ways, but this is what I got:

Using the definition of exclusive or:

$$\begin{align} r \overbrace{\oplus}^{\text{xor}} s \equiv \\( \sim r \land s) \lor (\sim s \land r) \equiv\\ \sim( \sim r \land s) \rightarrow (\sim s \land r) \equiv\\ (r \lor \sim s)\rightarrow ( \sim ( \sim (r \land \sim s))) \equiv \\ ( \sim r \rightarrow \sim s) \rightarrow ( \sim (\sim r \lor s)) \equiv\\ (\sim r \rightarrow \sim s) \rightarrow (\sim ( r \rightarrow s)) \tag{0} \end{align}$$

The known facts: $$\begin{align} A \rightarrow B \equiv (\sim A \lor B) \tag{1} \\ A \lor B \equiv \sim A \rightarrow B \tag{2} \\ \sim ( \sim A) \equiv A \tag{3} \end{align}$$

And De-Morgan's law:

$$ \begin{align}\sim (A \lor B) \equiv (\sim A) \land (\sim B) \tag{4} \\ \sim (A \land B) \equiv (\sim A) \lor (\sim B) \tag{5} \end{align} $$

Thus we have:

$$\begin{align} p \lor ( \sim q \land ( r \oplus s)) \equiv \\\sim p \rightarrow ( \sim q \land ( r \oplus s)) \equiv \tag{2} \\\sim p \rightarrow ( \sim ( \sim (\sim q \land (r \oplus s)))) \equiv \tag{3} \\\sim p \rightarrow ( \sim (q \lor \sim (r \oplus s))) \equiv \tag{4} \\\sim p \rightarrow ( \sim ( \sim q \rightarrow \sim (r \oplus s))) \end{align}$$

Substituting what we got for the xor above, we get:

$$ \equiv \tag{0}\\ \sim p \rightarrow ( \sim ( \sim q \rightarrow \sim ((\sim r \rightarrow \sim s) \rightarrow (\sim ( r \rightarrow s))))) $$

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