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Theorem sbc2or 3751
Description: The disjunction of two equivalences for class substitution does not require a class existence hypothesis. This theorem tells us that there are only 2 possibilities for [𝐴 / 𝑥]𝜑 behavior at proper classes, matching the sbc5 3770 (false) and sbc6 3773 (true) conclusions. This is interesting since dfsbcq 3744 and dfsbcq2 3745 (from which it is derived) do not appear to say anything obvious about proper class behavior. Note that this theorem does not tell us that it is always one or the other at proper classes; it could "flip" between false (the first disjunct) and true (the second disjunct) as a function of some other variable 𝑦 that 𝜑 or 𝐴 may contain. (Contributed by NM, 11-Oct-2004.) (Proof modification is discouraged.)
Assertion
Ref Expression
sbc2or (([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)) ∨ ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
Distinct variable group:   𝑥,𝐴
Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem sbc2or
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 dfsbcq2 3745 . . . 4 (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑[𝐴 / 𝑥]𝜑))
2 eqeq2 2774 . . . . . 6 (𝑦 = 𝐴 → (𝑥 = 𝑦𝑥 = 𝐴))
32anbi1d 643 . . . . 5 (𝑦 = 𝐴 → ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝐴𝜑)))
43exbidv 1954 . . . 4 (𝑦 = 𝐴 → (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
5 sb5 2311 . . . 4 ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑))
61, 4, 5vtoclbg 3522 . . 3 (𝐴 ∈ V → ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
76orcd 887 . 2 (𝐴 ∈ V → (([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)) ∨ ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑))))
8 pm5.15 1030 . . 3 (([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)) ∨ ([𝐴 / 𝑥]𝜑 ↔ ¬ ∃𝑥(𝑥 = 𝐴𝜑)))
9 vex 3457 . . . . . . . . . 10 𝑥 ∈ V
10 eleq1 2850 . . . . . . . . . 10 (𝑥 = 𝐴 → (𝑥 ∈ V ↔ 𝐴 ∈ V))
119, 10mpbii 236 . . . . . . . . 9 (𝑥 = 𝐴𝐴 ∈ V)
1211adantr 486 . . . . . . . 8 ((𝑥 = 𝐴𝜑) → 𝐴 ∈ V)
1312con3i 155 . . . . . . 7 𝐴 ∈ V → ¬ (𝑥 = 𝐴𝜑))
1413nexdv 1969 . . . . . 6 𝐴 ∈ V → ¬ ∃𝑥(𝑥 = 𝐴𝜑))
1511con3i 155 . . . . . . . 8 𝐴 ∈ V → ¬ 𝑥 = 𝐴)
1615pm2.21d 122 . . . . . . 7 𝐴 ∈ V → (𝑥 = 𝐴𝜑))
1716alrimiv 1960 . . . . . 6 𝐴 ∈ V → ∀𝑥(𝑥 = 𝐴𝜑))
1814, 172thd 268 . . . . 5 𝐴 ∈ V → (¬ ∃𝑥(𝑥 = 𝐴𝜑) ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
1918bibi2d 345 . . . 4 𝐴 ∈ V → (([𝐴 / 𝑥]𝜑 ↔ ¬ ∃𝑥(𝑥 = 𝐴𝜑)) ↔ ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑))))
2019orbi2d 929 . . 3 𝐴 ∈ V → ((([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)) ∨ ([𝐴 / 𝑥]𝜑 ↔ ¬ ∃𝑥(𝑥 = 𝐴𝜑))) ↔ (([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)) ∨ ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑)))))
218, 20mpbii 236 . 2 𝐴 ∈ V → (([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)) ∨ ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑))))
227, 21pm2.61i 184 1 (([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)) ∨ ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3  wi 4  wb 209  wa 401  wo 861  wal 1568   = wceq 1570  wex 1812  [wsb 2099  wcel 2145  Vcvv 3453  [wsbc 3742
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-10 2178  ax-12 2215  ax-ext 2734
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3455  df-sbc 3743
This theorem is used by: (None)
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