Theorem nssrex 4003

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 4002	. 2 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		df-rex 3090	. 2 ⊢ (∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	bitr4i 281	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)

Syntax hints:  ¬ wn 3   ↔ wb 209   ∧ wa 400  ∃wex 1809   ∈ wcel 2143  ∃wrex 3089   ⊆ wss 3906

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1810  df-rex 3090  df-ss 3923

This theorem is referenced by:  lnssplnglem  29051  dflring3  33765  dflring4  33766  mapssbi  45909