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Theorem bj-sbsb 31864
Description: Biconditional showing two possible (dual) definitions of substitution df-sb 1831 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 471 . . . 4 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
2 pm2.27 40 . . . . . 6 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → 𝜑))
32anc2li 577 . . . . 5 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑)))
43sps 1996 . . . 4 (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑)))
5 olc 397 . . . 4 ((𝑥 = 𝑦𝜑) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
61, 4, 5syl56 35 . . 3 (∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑))))
7 simpr 475 . . . 4 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → ∃𝑥(𝑥 = 𝑦𝜑))
8 equs5 2243 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
98biimpd 217 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦𝜑) → ∀𝑥(𝑥 = 𝑦𝜑)))
10 orc 398 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
117, 9, 10syl56 35 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑))))
126, 11pm2.61i 174 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
13 sp 1990 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
14 pm3.4 581 . . . 4 ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
1513, 14jaoi 392 . . 3 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
16 equs4 2181 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
17 19.8a 1988 . . . 4 ((𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
1816, 17jaoi 392 . . 3 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → ∃𝑥(𝑥 = 𝑦𝜑))
1915, 18jca 552 . 2 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
2012, 19impbii 197 1 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 194  wo 381  wa 382  wal 1472  wex 1694
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1700  ax-4 1713  ax-5 1793  ax-6 1838  ax-7 1885  ax-10 1966  ax-12 1983  ax-13 2137
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-ex 1695  df-nf 1699
This theorem is referenced by:  bj-dfsb2  31865
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