Theorem nssr 3099

Description: Negation of subclass relationship. One direction of Exercise 13 of [TakeutiZaring] p. 18. (Contributed by Jim Kingdon, 15-Jul-2018.)

Assertion

Ref	Expression
nssr	⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) → ¬ 𝐴 ⊆ 𝐵)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nssr

Step	Hyp	Ref	Expression
1		exanaliim 1590	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) → ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		dfss2 3028	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	sylnibr 640	1 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) → ¬ 𝐴 ⊆ 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103  ∀wal 1294  ∃wex 1433   ∈ wcel 1445   ⊆ wss 3013

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 582  ax-in2 583  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-11 1449  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077

This theorem depends on definitions:  df-bi 116  df-tru 1299  df-fal 1302  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-in 3019  df-ss 3026

This theorem is referenced by: (None)