Theorem nssrex 4001

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 4000	. 2 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		df-rex 3086	. 2 ⊢ (∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	bitr4i 280	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)

Syntax hints:  ¬ wn 3   ↔ wb 208   ∧ wa 399  ∃wex 1798   ∈ wcel 2141  ∃wrex 3085   ⊆ wss 3904

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1799  df-rex 3086  df-ss 3921

This theorem is referenced by:  dflring3  33654  dflring4  33655  mapssbi  45753