Theorem ssdisjdr 49430

Description: Subset preserves disjointness. Deduction form of ssdisj 4414. Alternatively this could be proved with ineqcom 4162 in tandem with ssdisjd 49429. (Contributed by Zhi Wang, 7-Sep-2024.)

Hypotheses

Ref	Expression
ssdisjd.1	⊢ (𝜑 → 𝐴 ⊆ 𝐵)
ssdisjdr.2	⊢ (𝜑 → (𝐶 ∩ 𝐵) = ∅)

Assertion

Ref	Expression
ssdisjdr	⊢ (𝜑 → (𝐶 ∩ 𝐴) = ∅)

Proof of Theorem ssdisjdr

Step	Hyp	Ref	Expression
1		ssdisjd.1	. . 3 ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
2		sslin 4194	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∩ 𝐴) ⊆ (𝐶 ∩ 𝐵))
3	1, 2	syl 17	. 2 ⊢ (𝜑 → (𝐶 ∩ 𝐴) ⊆ (𝐶 ∩ 𝐵))
4		ssdisjdr.2	. 2 ⊢ (𝜑 → (𝐶 ∩ 𝐵) = ∅)
5		sseq0 4357	. 2 ⊢ (((𝐶 ∩ 𝐴) ⊆ (𝐶 ∩ 𝐵) ∧ (𝐶 ∩ 𝐵) = ∅) → (𝐶 ∩ 𝐴) = ∅)
6	3, 4, 5	syl2anc 593	1 ⊢ (𝜑 → (𝐶 ∩ 𝐴) = ∅)

Syntax hints:   → wi 4   = wceq 1560   ∩ cin 3903   ⊆ wss 3904  ∅c0 4285

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1563  df-fal 1573  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3456  df-dif 3907  df-in 3911  df-ss 3921  df-nul 4286

This theorem is referenced by:  predisj  49432  seposep  49547