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Theorem nabbib 3064
Description: Not equivalent wff's correspond to not equal class abstractions. (Contributed by AV, 7-Apr-2019.) (Proof shortened by Wolf Lammen, 25-Nov-2019.) Definitial form. (Revised by Wolf Lammen, 5-Mar-2025.)
Assertion
Ref Expression
nabbib ({𝑥 ∣ 𝜑} ≠ {𝑥 ∣ 𝜓} ↔ ∃𝑥(𝜑 ↔ ¬ 𝜓))

Proof of Theorem nabbib
StepHypRef Expression
1 df-ne 2960 . 2 ({𝑥 ∣ 𝜑} ≠ {𝑥 ∣ 𝜓} ↔ ¬ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓})
2 exnal 1860 . . . 4 (∃𝑥 ¬ (𝜑 ↔ 𝜓) ↔ ¬ ∀𝑥(𝜑 ↔ 𝜓))
3 xor3 385 . . . . 5 (¬ (𝜑 ↔ 𝜓) ↔ (𝜑 ↔ ¬ 𝜓))
43exbii 1881 . . . 4 (∃𝑥 ¬ (𝜑 ↔ 𝜓) ↔ ∃𝑥(𝜑 ↔ ¬ 𝜓))
52, 4bitr3i 280 . . 3 (¬ ∀𝑥(𝜑 ↔ 𝜓) ↔ ∃𝑥(𝜑 ↔ ¬ 𝜓))
6 abbib 2832 . . 3 ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))
75, 6xchnxbir 336 . 2 (¬ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∃𝑥(𝜑 ↔ ¬ 𝜓))
81, 7bitri 278 1 ({𝑥 ∣ 𝜑} ≠ {𝑥 ∣ 𝜓} ↔ ∃𝑥(𝜑 ↔ ¬ 𝜓))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209  ∀wal 1568   = wceq 1570  ∃wex 1812  {cab 2741   ≠ wne 2959
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 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-ne 2960
This theorem is used by:  suppvalbr  8211  suppvalbrOLD  8212
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