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Theorem bj-sbsb 35016
Description: Biconditional showing two possible (dual) definitions of substitution df-sb 2072 not using dummy variables. (Contributed by BJ, 19-Mar-2021.)
Assertion
Ref Expression
bj-sbsb (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))

Proof of Theorem bj-sbsb
StepHypRef Expression
1 simpl 483 . . . 4 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
2 pm2.27 42 . . . . . 6 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → 𝜑))
32anc2li 556 . . . . 5 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑)))
43sps 2182 . . . 4 (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑)))
5 olc 865 . . . 4 ((𝑥 = 𝑦𝜑) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
61, 4, 5syl56 36 . . 3 (∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑))))
7 simpr 485 . . . 4 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → ∃𝑥(𝑥 = 𝑦𝜑))
8 equs5 2462 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
98biimpd 228 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦𝜑) → ∀𝑥(𝑥 = 𝑦𝜑)))
10 orc 864 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
117, 9, 10syl56 36 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑))))
126, 11pm2.61i 182 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
13 sp 2180 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
14 pm3.4 807 . . . 4 ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
1513, 14jaoi 854 . . 3 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
16 equs4 2418 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
17 19.8a 2178 . . . 4 ((𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
1816, 17jaoi 854 . . 3 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → ∃𝑥(𝑥 = 𝑦𝜑))
1915, 18jca 512 . 2 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
2012, 19impbii 208 1 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 205  wa 396  wo 844  wal 1540  wex 1786
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-10 2141  ax-12 2175  ax-13 2374
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-ex 1787  df-nf 1791
This theorem is referenced by:  bj-dfsb2  35017
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