Đáp án:

`@` Ta có:

`{:(P in AB sub (SAB)),(P in d sub (E,d)):}}` `=> P in (SAB)nn(E,d)` 

`@` Trong `(SAB):` gọi `I=SB nn EP`

`@` Ta có:

`{:(I in SB sub (SAB)),(I in EP sub (E,d)):}}`

`=> I in (SAB)nn(E,d)` (1)

`@` Ta có:

`{:(E in SA sub (SAB)),(E in (E,d)):}}`

`=> E in (SAB)nn(E,d)` (2)

`@`(1)(2)`=>IE = (SAB)nn(E,d)`

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