Theorem nssss 5410

Description: Negation of subclass relationship. Compare nss 4000. (Contributed by NM, 30-Jun-2004.) (Proof shortened by Andrew Salmon, 25-Jul-2011.)

Assertion

Ref	Expression
nssss	⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nssss

Step	Hyp	Ref	Expression
1		exanali 1861	. . 3 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵) ↔ ¬ ∀𝑥(𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵))
2		ssextss 5408	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵))
3	1, 2	xchbinxr 335	. 2 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵) ↔ ¬ 𝐴 ⊆ 𝐵)
4	3	bicomi 224	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1540  ∃wex 1781   ⊆ wss 3903

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5243  ax-pr 5379

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-v 3444  df-un 3908  df-ss 3920  df-pw 4558  df-sn 4583  df-pr 4585

This theorem is referenced by: (None)