Theorem ssdisjdr 48840

Description: Subset preserves disjointness. Deduction form of ssdisj 4405. Alternatively this could be proved with ineqcom 4155 in tandem with ssdisjd 48839. (Contributed by Zhi Wang, 7-Sep-2024.)

Hypotheses

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

Assertion

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

Proof of Theorem ssdisjdr

Step	Hyp	Ref	Expression
1		ssdisjd.1	. . 3 ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
2		sslin 4188	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∩ 𝐴) ⊆ (𝐶 ∩ 𝐵))
3	1, 2	syl 17	. 2 ⊢ (𝜑 → (𝐶 ∩ 𝐴) ⊆ (𝐶 ∩ 𝐵))
4		ssdisjdr.2	. 2 ⊢ (𝜑 → (𝐶 ∩ 𝐵) = ∅)
5		sseq0 4348	. 2 ⊢ (((𝐶 ∩ 𝐴) ⊆ (𝐶 ∩ 𝐵) ∧ (𝐶 ∩ 𝐵) = ∅) → (𝐶 ∩ 𝐴) = ∅)
6	3, 4, 5	syl2anc 584	1 ⊢ (𝜑 → (𝐶 ∩ 𝐴) = ∅)

Syntax hints:   → wi 4   = wceq 1541   ∩ cin 3896   ⊆ wss 3897  ∅c0 4278

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-v 3438  df-dif 3900  df-in 3904  df-ss 3914  df-nul 4279

This theorem is referenced by:  predisj  48842  seposep  48957