Theorem ssdisjd 47394

Description: Subset preserves disjointness. Deduction form of ssdisj 4458. (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 4234	. 2 ⊢ (𝜑 → (𝐴 ∩ 𝐶) ⊆ (𝐵 ∩ 𝐶))
3		ssdisjd.2	. 2 ⊢ (𝜑 → (𝐵 ∩ 𝐶) = ∅)
4		sseq0 4398	. 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 3946   ⊆ wss 3947  ∅c0 4321

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 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-fal 1555  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3477  df-dif 3950  df-in 3954  df-ss 3964  df-nul 4322

This theorem is referenced by:  predisj  47397  iccdisj2  47432  sepdisj  47459  seposep  47460