Metamath Proof Explorer


Theorem sscon

Description: Contraposition law for subsets. Exercise 15 of TakeutiZaring p. 22. (Contributed by NM, 22-Mar-1998)

Ref Expression
Assertion sscon ( 𝐴 ⊆ 𝐵 → ( 𝐶 ∖ 𝐵 ) ⊆ ( 𝐶 ∖ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 ssel ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
2 1 con3d ⊢ ( 𝐴 ⊆ 𝐵 → ( ¬ 𝑥 ∈ 𝐵 → ¬ 𝑥 ∈ 𝐴 ) )
3 2 anim2d ⊢ ( 𝐴 ⊆ 𝐵 → ( ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵 ) → ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐴 ) ) )
4 eldif ⊢ ( 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) ↔ ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵 ) )
5 eldif ⊢ ( 𝑥 ∈ ( 𝐶 ∖ 𝐴 ) ↔ ( 𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐴 ) )
6 3 4 5 3imtr4g ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑥 ∈ ( 𝐶 ∖ 𝐵 ) → 𝑥 ∈ ( 𝐶 ∖ 𝐴 ) ) )
7 6 ssrdv ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐶 ∖ 𝐵 ) ⊆ ( 𝐶 ∖ 𝐴 ) )