1)因
∈(A∩B)×C
a∈A∩B ∧ c∈C
(a∈A ∧ a∈B) ∧ c∈C
(a∈A ∧ c∈C) ∧ (a∈B ∧ c∈C)
∈A×C ∧ ∈B×C
∈(A×C)∩(B×C),
故得
(A∩B)×C = (A×C)∩(B×C)。
2)因
...
1)因
∈(A∩B)×C
a∈A∩B ∧ c∈C
(a∈A ∧ a∈B) ∧ c∈C
(a∈A ∧ c∈C) ∧ (a∈B ∧ c∈C)
∈A×C ∧ ∈B×C
∈(A×C)∩(B×C),
故得
(A∩B)×C = (A×C)∩(B×C)。
2)因
...