Football |A|=28 basketball |B|=29 volleyball | c | = 26 | a ∩ b | = 7 | b ∩ c | = 9 | a ∩ c | =1.

| A |+| B |+| C |-| A∩B |-| B∩C |-| A∩C |+| A∩B∩C | = | A∪B∪C | = 60

| a ∩ b ∩ c | = 60-28-29-26+7+9+11= 4, that is, there are four contestants in all three competitions.

2、

It's easy, but if you need a graph, there are four two-degree nodes and five leaves, so there are no three-degree nodes.

The calculation is also very simple, the number of leaves = total degree-number of nodes+1. Assume that the number of nodes in the third degree is x.

Namely: 2*4+x*3-4-x+ 1=5.

The solution is x=0.

∴ There are no 3D nodes.

3、B ∪~((~ A∪B)∩A)= B ∪~((~ A∪A)∩(B∪A))= B ∪~(B∪A)= B ∪( ~ B∪~ A)= B∪~ B∪~ A = U∪~ B = U

4、

It is proved that if x, y∈Z, then x ☉ y = x+y-2 ∈ z.

∴<； z,☉>； closed

For any x, y, z∈Z

(x☉y)☉z=(x+y-2)+z-2=x+(y+z-2)-2=x+(y☉z)-2=x☉(y☉z)

∴<； z,☉>； Combinable.

For any x∈Z

x☉2=x+2-2=x

2☉x=2+x-2=x

Yes.

X☉(4-x)=x+(4-x)-2=2= unitary

So 4-x is a unary of X.

To sum up: < z, ☉ > It's a group

5. Proof: (p → q) ∧ (q → p) <; = & gt(﹁p∨q)∧(﹁q∨p)<； = & gt((﹁p∨q)∧﹁q)∨((﹁p∨q)∧p)<； = & gt

((﹁p∧﹁q)∨(q∧﹁q))∨((﹁p∧p)∨(q∧p))<； = & gt(﹁p∧﹁q)∨(q∧p)<； = & gt

﹁(p∨q)∨(q∧p)<； = & gt(P∨Q)→(Q∧P)