Theorem dfss6 3998

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		df-ss 3993	. . 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 275	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ¬ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
4		exanali 1858	. 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 206   ∧ wa 395  ∀wal 1535  ∃wex 1777   ∈ wcel 2108   ⊆ wss 3976

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1778  df-ss 3993

This theorem is referenced by:  dfssr2  38455