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Theorem bj-sbsb 33159
Description: Biconditional showing two possible (dual) definitions of substitution df-sb 2050 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 468 . . . 4 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
2 pm2.27 42 . . . . . 6 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → 𝜑))
32anc2li 545 . . . . 5 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑)))
43sps 2209 . . . 4 (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑)))
5 olc 857 . . . 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 471 . . . 4 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → ∃𝑥(𝑥 = 𝑦𝜑))
8 equs5 2497 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
98biimpd 219 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦𝜑) → ∀𝑥(𝑥 = 𝑦𝜑)))
10 orc 856 . . . 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 176 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
13 sp 2207 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
14 pm3.4 811 . . . 4 ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
1513, 14jaoi 846 . . 3 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
16 equs4 2445 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
17 19.8a 2206 . . . 4 ((𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
1816, 17jaoi 846 . . 3 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → ∃𝑥(𝑥 = 𝑦𝜑))
1915, 18jca 501 . 2 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
2012, 19impbii 199 1 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 196  wa 382  wo 836  wal 1629  wex 1852
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-10 2174  ax-12 2203  ax-13 2408
This theorem depends on definitions:  df-bi 197  df-an 383  df-or 837  df-ex 1853  df-nf 1858
This theorem is referenced by:  bj-dfsb2  33160
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