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Theorem bj-sbsb 33399
Description: Biconditional showing two possible (dual) definitions of substitution df-sb 2012 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 476 . . . 4 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
2 pm2.27 42 . . . . . 6 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → 𝜑))
32anc2li 551 . . . . 5 (𝑥 = 𝑦 → ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑)))
43sps 2169 . . . 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 479 . . . 4 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → ∃𝑥(𝑥 = 𝑦𝜑))
8 equs5 2426 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜑)))
98biimpd 221 . . . 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 177 . 2 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) → (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
13 sp 2167 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
14 pm3.4 800 . . . 4 ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
1513, 14jaoi 846 . . 3 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → (𝑥 = 𝑦𝜑))
16 equs4 2381 . . . 4 (∀𝑥(𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
17 19.8a 2166 . . . 4 ((𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
1816, 17jaoi 846 . . 3 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → ∃𝑥(𝑥 = 𝑦𝜑))
1915, 18jca 507 . 2 ((∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)) → ((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
2012, 19impbii 201 1 (((𝑥 = 𝑦𝜑) ∧ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ (∀𝑥(𝑥 = 𝑦𝜑) ∨ (𝑥 = 𝑦𝜑)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 198  wa 386  wo 836  wal 1599  wex 1823
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-10 2135  ax-12 2163  ax-13 2334
This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-ex 1824  df-nf 1828
This theorem is referenced by:  bj-dfsb2  33400
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