Theorem rexbidar 40785

Description: More general form of rexbida 3320. (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

Step	Hyp	Ref	Expression
1		ralbidar.1	. . . . 5 ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 𝜑)
2		ralbidar.2	. . . . . . 7 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
3	2	ex 415	. . . . . 6 ⊢ (𝜑 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
4	3	ralimi 3162	. . . . 5 ⊢ (∀𝑥 ∈ 𝐴 𝜑 → ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
5	1, 4	syl 17	. . . 4 ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
6		df-ral 3145	. . . 4 ⊢ (∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)) ↔ ∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))))
7	5, 6	sylib 220	. . 3 ⊢ (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))))
8		pm2.43 56	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
9	8	pm5.32d 579	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)))
10	9	alimi 1812	. . 3 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)))
11		exbi 1847	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)) → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓) ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒)))
12	7, 10, 11	3syl 18	. 2 ⊢ (𝜑 → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓) ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒)))
13		df-rex 3146	. 2 ⊢ (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓))
14		df-rex 3146	. 2 ⊢ (∃𝑥 ∈ 𝐴 𝜒 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒))
15	12, 13, 14	3bitr4g 316	1 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398  ∀wal 1535  ∃wex 1780   ∈ wcel 2114  ∀wral 3140  ∃wrex 3141

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1781  df-ral 3145  df-rex 3146

This theorem is referenced by: (None)