Theorem ssdisjd 47956

Description: Subset preserves disjointness. Deduction form of ssdisj 4463. (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 4238	. 2 ⊢ (𝜑 → (𝐴 ∩ 𝐶) ⊆ (𝐵 ∩ 𝐶))
3		ssdisjd.2	. 2 ⊢ (𝜑 → (𝐵 ∩ 𝐶) = ∅)
4		sseq0 4403	. 2 ⊢ (((𝐴 ∩ 𝐶) ⊆ (𝐵 ∩ 𝐶) ∧ (𝐵 ∩ 𝐶) = ∅) → (𝐴 ∩ 𝐶) = ∅)
5	2, 3, 4	syl2anc 582	1 ⊢ (𝜑 → (𝐴 ∩ 𝐶) = ∅)

Syntax hints:   → wi 4   = wceq 1533   ∩ cin 3948   ⊆ wss 3949  ∅c0 4326

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1536  df-fal 1546  df-ex 1774  df-sb 2060  df-clab 2706  df-cleq 2720  df-clel 2806  df-v 3475  df-dif 3952  df-in 3956  df-ss 3966  df-nul 4327

This theorem is referenced by:  predisj  47959  iccdisj2  47994  sepdisj  48021  seposep  48022