Theorem rexbidar 37671

Description: More general form of rexbida 3029. (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 449	. . . . . 6 ⊢ (𝜑 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
4	3	ralimi 2936	. . . . 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 2901	. . . 4 ⊢ (∀𝑥 ∈ 𝐴 (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)) ↔ ∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))))
7	5, 6	sylib 207	. . 3 ⊢ (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))))
8		pm2.43 54	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒)))
9	8	pm5.32d 669	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)))
10	9	alimi 1730	. . 3 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝜓 ↔ 𝜒))) → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐴 ∧ 𝜒)))
11		exbi 1762	. . 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 2902	. 2 ⊢ (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜓))
14		df-rex 2902	. 2 ⊢ (∃𝑥 ∈ 𝐴 𝜒 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜒))
15	12, 13, 14	3bitr4g 302	1 ⊢ (𝜑 → (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑥 ∈ 𝐴 𝜒))

Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383  ∀wal 1473  ∃wex 1695   ∈ wcel 1977  ∀wral 2896  ∃wrex 2897

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728

This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696  df-ral 2901  df-rex 2902

This theorem is referenced by: (None)