Theorem ssdisjd 46153

Description: Subset preserves disjointness. Deduction form of ssdisj 4393. (Contributed by Zhi Wang, 7-Sep-2024.)

Hypotheses

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

Assertion

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

Proof of Theorem ssdisjd

Step	Hyp	Ref	Expression
1		ssdisjd.1	. . 3 ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
2	1	ssrind 4169	. 2 ⊢ (𝜑 → (𝐴 ∩ 𝐶) ⊆ (𝐵 ∩ 𝐶))
3		ssdisjd.2	. 2 ⊢ (𝜑 → (𝐵 ∩ 𝐶) = ∅)
4		sseq0 4333	. 2 ⊢ (((𝐴 ∩ 𝐶) ⊆ (𝐵 ∩ 𝐶) ∧ (𝐵 ∩ 𝐶) = ∅) → (𝐴 ∩ 𝐶) = ∅)
5	2, 3, 4	syl2anc 584	1 ⊢ (𝜑 → (𝐴 ∩ 𝐶) = ∅)

Syntax hints:   → wi 4   = wceq 1539   ∩ cin 3886   ⊆ wss 3887  ∅c0 4256

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-dif 3890  df-in 3894  df-ss 3904  df-nul 4257

This theorem is referenced by:  predisj  46156  iccdisj2  46191  sepdisj  46218  seposep  46219