Theorem nssss 5419

Description: Negation of subclass relationship. Compare nss 3998. (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 1878	. . 3 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵) ↔ ¬ ∀𝑥(𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵))
2		ssextss 5417	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵))
3	1, 2	xchbinxr 337	. 2 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵) ↔ ¬ 𝐴 ⊆ 𝐵)
4	3	bicomi 226	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 399  ∀wal 1557  ∃wex 1798   ⊆ wss 3902

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733  ax-sep 5243  ax-pr 5387

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3455  df-un 3907  df-ss 3919  df-pw 4554  df-sn 4580  df-pr 4582

This theorem is referenced by: (None)