Theorem ssdisjdr 49646

Description: Subset preserves disjointness. Deduction form of ssdisj 4420. Alternatively this could be proved with ineqcom 4163 in tandem with ssdisjd 49645. (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 4195	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∩ 𝐴) ⊆ (𝐶 ∩ 𝐵))
3	1, 2	syl 18	. 2 ⊢ (𝜑 → (𝐶 ∩ 𝐴) ⊆ (𝐶 ∩ 𝐵))
4		ssdisjdr.2	. 2 ⊢ (𝜑 → (𝐶 ∩ 𝐵) = ∅)
5		sseq0 4361	. 2 ⊢ (((𝐶 ∩ 𝐴) ⊆ (𝐶 ∩ 𝐵) ∧ (𝐶 ∩ 𝐵) = ∅) → (𝐶 ∩ 𝐴) = ∅)
6	3, 4, 5	syl2anc 596	1 ⊢ (𝜑 → (𝐶 ∩ 𝐴) = ∅)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∩ cin 3905   ⊆ wss 3906  ∅c0 4286

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-in 3913  df-ss 3923  df-nul 4287

This theorem is used by:  predisj  49648  seposep  49763