Theorem ssdisjdr 50058

Description: Subset preserves disjointness. Deduction form of ssdisj 4415. Alternatively this could be proved with ineqcom 4109 in tandem with ssdisjd 50057. (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 4141	. . 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 4356	. 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 1569   ⊆ wss 3899   ∩ cin 4091  ∅c0 4280

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3416  df-v 3455  df-ss 3900  df-dif 4049  df-in 4093  df-nul 4281

This theorem is used by:  predisj  50060  seposep  50179