Theorem dfss6 3910

Description: Alternate definition of subclass relationship. (Contributed by RP, 16-Apr-2020.)

Assertion

Ref	Expression
dfss6	⊢ (𝐴 ⊆ 𝐵 ↔ ¬ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem dfss6

Step	Hyp	Ref	Expression
1		dfss2 3907	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		notnotb 315	. . 3 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ ¬ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	bitri 274	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ¬ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
4		exanali 1862	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
5	3, 4	xchbinxr 335	1 ⊢ (𝐴 ⊆ 𝐵 ↔ ¬ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 396  ∀wal 1537  ∃wex 1782   ∈ wcel 2106   ⊆ wss 3887

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 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-in 3894  df-ss 3904

This theorem is referenced by:  dfssr2  36617