Theorem nssrex 42525

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 3979	. 2 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		df-rex 3069	. 2 ⊢ (∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	bitr4i 277	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)

Syntax hints:  ¬ wn 3   ↔ wb 205   ∧ wa 395  ∃wex 1783   ∈ wcel 2108  ∃wrex 3064   ⊆ wss 3883

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-rex 3069  df-v 3424  df-in 3890  df-ss 3900

This theorem is referenced by:  mapssbi  42642