Odpowiedź :
(p∨q)=>p
1 1 1 1 1
1 1 0 1 1
0 1 1 1 0
0 0 0 1 0
~[(p∨q)=>(p∧q)]
0 1 1 1 1 1 1
1 1 0 0 1 0 0
1 0 1 0 0 0 1
0 0 0 1 0 0 0
1 1 1 1 1
1 1 0 1 1
0 1 1 1 0
0 0 0 1 0
~[(p∨q)=>(p∧q)]
0 1 1 1 1 1 1
1 1 0 0 1 0 0
1 0 1 0 0 0 1
0 0 0 1 0 0 0