Metamath Proof Explorer


Theorem ssconb

Description: Contraposition law for subsets. (Contributed by NM, 22-Mar-1998)

Ref Expression
Assertion ssconb ( ( 𝐴𝐶𝐵𝐶 ) → ( 𝐴 ⊆ ( 𝐶𝐵 ) ↔ 𝐵 ⊆ ( 𝐶𝐴 ) ) )

Proof

Step Hyp Ref Expression
1 ssel ( 𝐴𝐶 → ( 𝑥𝐴𝑥𝐶 ) )
2 ssel ( 𝐵𝐶 → ( 𝑥𝐵𝑥𝐶 ) )
3 pm5.1 ( ( ( 𝑥𝐴𝑥𝐶 ) ∧ ( 𝑥𝐵𝑥𝐶 ) ) → ( ( 𝑥𝐴𝑥𝐶 ) ↔ ( 𝑥𝐵𝑥𝐶 ) ) )
4 1 2 3 syl2an ( ( 𝐴𝐶𝐵𝐶 ) → ( ( 𝑥𝐴𝑥𝐶 ) ↔ ( 𝑥𝐵𝑥𝐶 ) ) )
5 con2b ( ( 𝑥𝐴 → ¬ 𝑥𝐵 ) ↔ ( 𝑥𝐵 → ¬ 𝑥𝐴 ) )
6 5 a1i ( ( 𝐴𝐶𝐵𝐶 ) → ( ( 𝑥𝐴 → ¬ 𝑥𝐵 ) ↔ ( 𝑥𝐵 → ¬ 𝑥𝐴 ) ) )
7 4 6 anbi12d ( ( 𝐴𝐶𝐵𝐶 ) → ( ( ( 𝑥𝐴𝑥𝐶 ) ∧ ( 𝑥𝐴 → ¬ 𝑥𝐵 ) ) ↔ ( ( 𝑥𝐵𝑥𝐶 ) ∧ ( 𝑥𝐵 → ¬ 𝑥𝐴 ) ) ) )
8 jcab ( ( 𝑥𝐴 → ( 𝑥𝐶 ∧ ¬ 𝑥𝐵 ) ) ↔ ( ( 𝑥𝐴𝑥𝐶 ) ∧ ( 𝑥𝐴 → ¬ 𝑥𝐵 ) ) )
9 jcab ( ( 𝑥𝐵 → ( 𝑥𝐶 ∧ ¬ 𝑥𝐴 ) ) ↔ ( ( 𝑥𝐵𝑥𝐶 ) ∧ ( 𝑥𝐵 → ¬ 𝑥𝐴 ) ) )
10 7 8 9 3bitr4g ( ( 𝐴𝐶𝐵𝐶 ) → ( ( 𝑥𝐴 → ( 𝑥𝐶 ∧ ¬ 𝑥𝐵 ) ) ↔ ( 𝑥𝐵 → ( 𝑥𝐶 ∧ ¬ 𝑥𝐴 ) ) ) )
11 eldif ( 𝑥 ∈ ( 𝐶𝐵 ) ↔ ( 𝑥𝐶 ∧ ¬ 𝑥𝐵 ) )
12 11 imbi2i ( ( 𝑥𝐴𝑥 ∈ ( 𝐶𝐵 ) ) ↔ ( 𝑥𝐴 → ( 𝑥𝐶 ∧ ¬ 𝑥𝐵 ) ) )
13 eldif ( 𝑥 ∈ ( 𝐶𝐴 ) ↔ ( 𝑥𝐶 ∧ ¬ 𝑥𝐴 ) )
14 13 imbi2i ( ( 𝑥𝐵𝑥 ∈ ( 𝐶𝐴 ) ) ↔ ( 𝑥𝐵 → ( 𝑥𝐶 ∧ ¬ 𝑥𝐴 ) ) )
15 10 12 14 3bitr4g ( ( 𝐴𝐶𝐵𝐶 ) → ( ( 𝑥𝐴𝑥 ∈ ( 𝐶𝐵 ) ) ↔ ( 𝑥𝐵𝑥 ∈ ( 𝐶𝐴 ) ) ) )
16 15 albidv ( ( 𝐴𝐶𝐵𝐶 ) → ( ∀ 𝑥 ( 𝑥𝐴𝑥 ∈ ( 𝐶𝐵 ) ) ↔ ∀ 𝑥 ( 𝑥𝐵𝑥 ∈ ( 𝐶𝐴 ) ) ) )
17 dfss2 ( 𝐴 ⊆ ( 𝐶𝐵 ) ↔ ∀ 𝑥 ( 𝑥𝐴𝑥 ∈ ( 𝐶𝐵 ) ) )
18 dfss2 ( 𝐵 ⊆ ( 𝐶𝐴 ) ↔ ∀ 𝑥 ( 𝑥𝐵𝑥 ∈ ( 𝐶𝐴 ) ) )
19 16 17 18 3bitr4g ( ( 𝐴𝐶𝐵𝐶 ) → ( 𝐴 ⊆ ( 𝐶𝐵 ) ↔ 𝐵 ⊆ ( 𝐶𝐴 ) ) )