Metamath Proof Explorer


Theorem disjss

Description: Subclass theorem for disjoints. (Contributed by Peter Mazsa, 28-Oct-2020) (Revised by Peter Mazsa, 22-Sep-2021)

Ref Expression
Assertion disjss ( 𝐴 ⊆ 𝐵 → ( Disj 𝐵 → Disj 𝐴 ) )

Proof

Step Hyp Ref Expression
1 cnvss ⊢ ( 𝐴 ⊆ 𝐵 → ◡ 𝐴 ⊆ ◡ 𝐵 )
2 funALTVss ⊢ ( ◡ 𝐴 ⊆ ◡ 𝐵 → ( FunALTV ◡ 𝐵 → FunALTV ◡ 𝐴 ) )
3 1 2 syl ⊢ ( 𝐴 ⊆ 𝐵 → ( FunALTV ◡ 𝐵 → FunALTV ◡ 𝐴 ) )
4 relss ⊢ ( 𝐴 ⊆ 𝐵 → ( Rel 𝐵 → Rel 𝐴 ) )
5 3 4 anim12d ⊢ ( 𝐴 ⊆ 𝐵 → ( ( FunALTV ◡ 𝐵 ∧ Rel 𝐵 ) → ( FunALTV ◡ 𝐴 ∧ Rel 𝐴 ) ) )
6 dfdisjALTV ⊢ ( Disj 𝐵 ↔ ( FunALTV ◡ 𝐵 ∧ Rel 𝐵 ) )
7 dfdisjALTV ⊢ ( Disj 𝐴 ↔ ( FunALTV ◡ 𝐴 ∧ Rel 𝐴 ) )
8 5 6 7 3imtr4g ⊢ ( 𝐴 ⊆ 𝐵 → ( Disj 𝐵 → Disj 𝐴 ) )