Theorem dfss6 3889

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 3886	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		notnotb 318	. . 3 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) ↔ ¬ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	bitri 278	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ¬ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
4		exanali 1867	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
5	3, 4	xchbinxr 338	1 ⊢ (𝐴 ⊆ 𝐵 ↔ ¬ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1541  ∃wex 1787   ∈ wcel 2110   ⊆ wss 3866

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-8 2112  ax-9 2120  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1546  df-ex 1788  df-sb 2071  df-clab 2715  df-cleq 2729  df-clel 2816  df-v 3410  df-in 3873  df-ss 3883

This theorem is referenced by:  dfssr2  36354