Theorem nssrex 44990

Description: Negation of subclass relationship. (Contributed by Glauco Siliprandi, 3-Mar-2021.)

Assertion

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

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nssrex

Step	Hyp	Ref	Expression
1		nss 4073	. 2 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		df-rex 3077	. 2 ⊢ (∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	bitr4i 278	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)

Syntax hints:  ¬ wn 3   ↔ wb 206   ∧ wa 395  ∃wex 1777   ∈ wcel 2108  ∃wrex 3076   ⊆ 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-rex 3077  df-ss 3993

This theorem is referenced by:  mapssbi  45122