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Theorem bj-subst 36046
Description: Proof of sbalex 2227 from core axioms, ax-10 2129 (modal5), and bj-ax12 36042. (Contributed by BJ, 29-Dec-2020.) (Proof modification is discouraged.)
Assertion
Ref Expression
bj-subst (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜑))
Distinct variable group:   𝑥,𝑦
Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem bj-subst
StepHypRef Expression
1 bj-ax12 36042 . . . 4 𝑥(𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))
2 pm3.31 449 . . . . 5 ((𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑))) → ((𝑥 = 𝑦𝜑) → ∀𝑥(𝑥 = 𝑦𝜑)))
32aleximi 1826 . . . 4 (∀𝑥(𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑))) → (∃𝑥(𝑥 = 𝑦𝜑) → ∃𝑥𝑥(𝑥 = 𝑦𝜑)))
41, 3ax-mp 5 . . 3 (∃𝑥(𝑥 = 𝑦𝜑) → ∃𝑥𝑥(𝑥 = 𝑦𝜑))
5 hbe1a 2132 . . 3 (∃𝑥𝑥(𝑥 = 𝑦𝜑) → ∀𝑥(𝑥 = 𝑦𝜑))
64, 5syl 17 . 2 (∃𝑥(𝑥 = 𝑦𝜑) → ∀𝑥(𝑥 = 𝑦𝜑))
7 equs4v 1995 . 2 (∀𝑥(𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
86, 7impbii 208 1 (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 395  wal 1531  wex 1773
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-10 2129  ax-12 2163
This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774
This theorem is referenced by: (None)
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