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Theorem nalsetOLD 5269
Description: Obsolete version of nalset 5268 as of 12-Apr-2026. (Contributed by NM, 23-Aug-1993.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
nalsetOLD ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥
Distinct variable group:   𝑥,𝑦

Proof of Theorem nalsetOLD
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 alexn 1878 . 2 (∀𝑥∃𝑦 ¬ 𝑦 ∈ 𝑥 ↔ ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥)
2 ax-sep 5249 . . 3 ∃𝑦∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧))
3 elequ1 2152 . . . . . 6 (𝑧 = 𝑦 → (𝑧 ∈ 𝑦 ↔ 𝑦 ∈ 𝑦))
4 elequ1 2152 . . . . . . 7 (𝑧 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑦 ∈ 𝑥))
5 elequ1 2152 . . . . . . . . 9 (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑧))
6 elequ2 2160 . . . . . . . . 9 (𝑧 = 𝑦 → (𝑦 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
75, 6bitrd 282 . . . . . . . 8 (𝑧 = 𝑦 → (𝑧 ∈ 𝑧 ↔ 𝑦 ∈ 𝑦))
87notbid 321 . . . . . . 7 (𝑧 = 𝑦 → (¬ 𝑧 ∈ 𝑧 ↔ ¬ 𝑦 ∈ 𝑦))
94, 8anbi12d 644 . . . . . 6 (𝑧 = 𝑦 → ((𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧) ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
103, 9bibi12d 348 . . . . 5 (𝑧 = 𝑦 → ((𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) ↔ (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦))))
1110spvv 2021 . . . 4 (∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) → (𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)))
12 pclem6 1043 . . . 4 ((𝑦 ∈ 𝑦 ↔ (𝑦 ∈ 𝑥 ∧ ¬ 𝑦 ∈ 𝑦)) → ¬ 𝑦 ∈ 𝑥)
1311, 12syl 18 . . 3 (∀𝑧(𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑥 ∧ ¬ 𝑧 ∈ 𝑧)) → ¬ 𝑦 ∈ 𝑥)
142, 13eximii 1870 . 2 ∃𝑦 ¬ 𝑦 ∈ 𝑥
151, 14mpgbi 1831 1 ¬ ∃𝑥∀𝑦 𝑦 ∈ 𝑥
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812
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 2147  ax-9 2155  ax-sep 5249
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813
This theorem is used by: (None)
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