let p be Sophie_Germain Prime; :: thesis: ( not p > 2 or p mod 4 = 1 or p mod 4 = 3 )
assume A1: p > 2 ; :: thesis: ( p mod 4 = 1 or p mod 4 = 3 )
set k = p mod 4;
( 0 <= p mod 4 & p mod 4 < 3 + 1 ) by INT_3:9;
then A2: ( 0 <= p mod 4 & p mod 4 <= 0 + 3 ) by NAT_1:13;

now

assume A3: ( p mod 4 = 0 or p mod 4 = 2 ) ; :: thesis: contradiction

now

per cases ( p mod 4 = 0 or p mod 4 = 2 ) by A3;

suppose p mod 4 = 0 ; :: thesis: contradiction

then 4 divides p by INT_1:89;
hence contradiction by INT_2:def 5, INT_2:46; :: thesis: verum

end;

suppose p mod 4 = 2 ; :: thesis: contradiction

then p = ((p div 4) * 4) + 2 by INT_1:86
.= (((p div 4) * 2) + 1) * 2 ;
then 2 divides p by INT_1:def 9;
hence contradiction by INT_2:def 5, A1; :: thesis: verum

end;

end;

end;

hence contradiction ; :: thesis: verum

end;

hence ( p mod 4 = 1 or p mod 4 = 3 ) by A2, NAT_1:28; :: thesis: verum