Theorem dfss6 3928

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 3923	. . 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 1892	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
5	3, 4	xchbinxr 338	1 ⊢ (𝐴 ⊆ 𝐵 ↔ ¬ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812   ∈ wcel 2146   ⊆ wss 3906

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-ss 3923

This theorem is used by:  dfssr2  39288