Theorem nssrex 4005

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 4004	. 2 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
2		df-rex 3093	. 2 ⊢ (∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
3	1, 2	bitr4i 281	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 401  ∃wex 1812   ∈ wcel 2146  ∃wrex 3092   ⊆ wss 3908

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-rex 3093  df-ss 3925

This theorem is used by:  lnssplnglem  29110  dflring3  33818  dflring4  33819  mapssbi  45970