Theorem nssss 5457

Description: Negation of subclass relationship. Compare nss 4044. (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 1855	. . 3 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵) ↔ ¬ ∀𝑥(𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵))
2		ssextss 5455	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵))
3	1, 2	xchbinxr 335	. 2 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵) ↔ ¬ 𝐴 ⊆ 𝐵)
4	3	bicomi 223	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 395  ∀wal 1532  ∃wex 1774   ⊆ wss 3947

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2699  ax-sep 5299  ax-pr 5429

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-ex 1775  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-v 3473  df-un 3952  df-in 3954  df-ss 3964  df-pw 4605  df-sn 4630  df-pr 4632

This theorem is referenced by: (None)