Theorem ssdisjd 45732

Description: Subset preserves disjointness. Deduction form of ssdisj 4359. (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 4136	. 2 ⊢ (𝜑 → (𝐴 ∩ 𝐶) ⊆ (𝐵 ∩ 𝐶))
3		ssdisjd.2	. 2 ⊢ (𝜑 → (𝐵 ∩ 𝐶) = ∅)
4		sseq0 4298	. 2 ⊢ (((𝐴 ∩ 𝐶) ⊆ (𝐵 ∩ 𝐶) ∧ (𝐵 ∩ 𝐶) = ∅) → (𝐴 ∩ 𝐶) = ∅)
5	2, 3, 4	syl2anc 587	1 ⊢ (𝜑 → (𝐴 ∩ 𝐶) = ∅)

Syntax hints:   → wi 4   = wceq 1542   ∩ cin 3852   ⊆ wss 3853  ∅c0 4221

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-ext 2711

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1545  df-fal 1555  df-ex 1787  df-sb 2075  df-clab 2718  df-cleq 2731  df-clel 2812  df-v 3402  df-dif 3856  df-in 3860  df-ss 3870  df-nul 4222

This theorem is referenced by:  predisj  45735  iccdisj2  45761  sepdisj  45787  seposep  45788