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Theorem rexbidar 45429
Description: More general form of rexbida 3274. (Contributed by Andrew Salmon, 25-Jul-2011.)
Hypotheses
Ref Expression
ralbidar.1 (𝜑 → ∀𝑥 ∈ 𝐴 𝜑)
ralbidar.2 ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
Assertion
Ref Expression
rexbidar (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))

Proof of Theorem rexbidar
StepHypRef Expression
1 ralbidar.1 . . . . 5 (𝜑 → ∀𝑥 ∈ 𝐴 𝜑)
2 ralbidar.2 . . . . . . 7 ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
32ex 418 . . . . . 6 (𝜑 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
43ralimi 3099 . . . . 5 (∀𝑥 ∈ 𝐴 𝜑 → ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
51, 4syl 18 . . . 4 (𝜑 → ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
6 df-ral 3077 . . . 4 (∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)) ↔ ∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))))
75, 6sylib 221 . . 3 (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))))
8 pm2.43 57 . . . . 5 ((𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
98pm5.32d 588 . . . 4 ((𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)))
109alimi 1843 . . 3 (∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)))
11 exbi 1879 . . 3 (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)) → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓) ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒)))
127, 10, 113syl 19 . 2 (𝜑 → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓) ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒)))
13 df-rex 3087 . 2 (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓))
14 df-rex 3087 . 2 (∃𝑥 ∈ 𝐴 𝜒 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒))
1512, 13, 143bitr4g 317 1 (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567  ∃wex 1811   ∈ wcel 2144  ∀wral 3076  ∃wrex 3086
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-ral 3077  df-rex 3087
This theorem is used by: (None)
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