e.g:Prove :If $A⊆B$ and $C⊆D$,then $A∩ C ⊆ B ∩ D$
Direct Proof:
1.$∀x(x∈A→x∈B)$(given)
2.$∀x(x∈C→x∈D)$(given)
3.$x∈A$
4.$x∈C$
5.$x∈B$(1,3 Modus ponens)
6.$x∈D$(2,4 Modus ponens)
7.$∀x(x∈A∧x∈C)→(x∈B∧x∈D)$ (conditon prove+Universal generation)
But it seem don't cover the all cases of whether x belong to A,B,C,D, so i will use use proof by cases but it is time consuming,as the following:
for $∀x((x∈A→x∈B)∧(x∈C→x∈D))→((x∈A∧x∈C)→(x∈B∧x∈D)))$
if $x∈A,x∈B,x∈C,x∈D$ then...
if $x∈A,x∈B,x∈C,x∉D$ then...
if $x∈A,x∈B,x∈C,x∈D$ then...
if $x∈A,x∈B,x∉C,x∈D$ then...
if $x∈A,x∈B,x∉C,x∉D$ then...
if $x∈A,x∉B,x∈C,x∈D$ then...
if $x∈A,x∉B,x∉C,x∈D$ then...
if $x∈A,x∉B,x∉C,x∉D$ then...
. . .
And seem that it is correct in every time,so i guess direct proof cover all the cases too?But how?