Theorem ssdisjd 49298

Description: Subset preserves disjointness. Deduction form of ssdisj 4401. (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 4185	. 2 ⊢ (𝜑 → (𝐴 ∩ 𝐶) ⊆ (𝐵 ∩ 𝐶))
3		ssdisjd.2	. 2 ⊢ (𝜑 → (𝐵 ∩ 𝐶) = ∅)
4		sseq0 4344	. 2 ⊢ (((𝐴 ∩ 𝐶) ⊆ (𝐵 ∩ 𝐶) ∧ (𝐵 ∩ 𝐶) = ∅) → (𝐴 ∩ 𝐶) = ∅)
5	2, 3, 4	syl2anc 585	1 ⊢ (𝜑 → (𝐴 ∩ 𝐶) = ∅)

Syntax hints:   → wi 4   = wceq 1542   ∩ cin 3889   ⊆ wss 3890  ∅c0 4274

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-v 3432  df-dif 3893  df-in 3897  df-ss 3907  df-nul 4275

This theorem is referenced by:  predisj  49301  iccdisj2  49387  sepdisj  49415  seposep  49416