Theorem nssss 5441

Description: Negation of subclass relationship. Compare nss 4004. (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 1892	. . 3 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵) ↔ ¬ ∀𝑥(𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵))
2		ssextss 5439	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵))
3	1, 2	xchbinxr 338	. 2 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵) ↔ ¬ 𝐴 ⊆ 𝐵)
4	3	bicomi 227	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ⊆ 𝐴 ∧ ¬ 𝑥 ⊆ 𝐵))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812   ⊆ 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  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738  ax-sep 5262  ax-pr 5409

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-v 3460  df-un 3913  df-ss 3925  df-pw 4569  df-sn 4595  df-pr 4597

This theorem is used by: (None)