P(x) : x is politician
Q(x) : x is crooked
Stmt : if everyone is politician then somebody is crooked
s2 can be directly inferred from stmt
∀(x) P(x) --> ∃(x) Q(x)
So s2 is true.
s2 :
= ∀(x) P(x) --> ∃(x) Q(x)
= ~( ∀(x)P(x) ) V ( ∃(x)Q(x) ) [Applied P->Q == ~P V Q]...equation 1
= ( ∃(x)~P(x)) V ( ∃(x)Q(x) ) [Applied ~[∀(x)P(x)] == ∃(x)~P(x) ]
= ∃(x) [ ~P(x) V Q(x) ]
= ∃(x) [ P(x) --> Q(x) ] ....this is S1
Now from equation 1 back
s2 :
= ~( ∀(x)P(x) ) V ( ∃(x)Q(x) )
= ~( ∀(x)P(x) ) V ( ~~∃(x)Q(x) ) [Applied ~~P(x) == P(x) ]
= ~( ∀(x)P(x) ) V ( ~∀(x)~Q(x) ) [Applied ~[∃(x)P(x)] == ∀(x)~P(x) ]
= ~ [ ( ∀(x)P(x) ) ^ ∀(x)~Q(x) ] [Applied ~ (P ^ Q) == ~P V ~Q]
= ~∀(x) [ P(x) ^ ~Q(x) ] ---> this is s3
Hence all options s1,s2 and s3 are equivalent.