Theorem dffr2ALT 5625

Description: Alternate proof of dffr2 5624, which avoids ax-8 2145 but requires ax-10 2176, ax-11 2192, ax-12 2213. (Contributed by NM, 17-Feb-2004.) (Proof shortened by Andrew Salmon, 27-Aug-2011.) (Proof shortened by Mario Carneiro, 23-Jun-2015.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
dffr2ALT	⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 {𝑧 ∈ 𝑥 ∣ 𝑧𝑅𝑦} = ∅))

Distinct variable groups:   𝑥,𝑦,𝑧,𝐴   𝑥,𝑅,𝑦,𝑧

Proof of Theorem dffr2ALT

Step	Hyp	Ref	Expression
1		df-fr 5616	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
2		rabeq0 4346	. . . . 5 ⊢ ({𝑧 ∈ 𝑥 ∣ 𝑧𝑅𝑦} = ∅ ↔ ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)
3	2	rexbii 3112	. . . 4 ⊢ (∃𝑦 ∈ 𝑥 {𝑧 ∈ 𝑥 ∣ 𝑧𝑅𝑦} = ∅ ↔ ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)
4	3	imbi2i 339	. . 3 ⊢ (((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 {𝑧 ∈ 𝑥 ∣ 𝑧𝑅𝑦} = ∅) ↔ ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
5	4	albii 1849	. 2 ⊢ (∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 {𝑧 ∈ 𝑥 ∣ 𝑧𝑅𝑦} = ∅) ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
6	1, 5	bitr4i 281	1 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 {𝑧 ∈ 𝑥 ∣ 𝑧𝑅𝑦} = ∅))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400  ∀wal 1568   = wceq 1570   ≠ wne 2958  ∀wral 3079  ∃wrex 3089  {crab 3416   ⊆ wss 3906  ∅c0 4287   class class class wbr 5110   Fr wfr 5613

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-fal 1583  df-ex 1810  df-nf 1814  df-sb 2097  df-clab 2742  df-cleq 2755  df-ral 3080  df-rex 3090  df-rab 3417  df-dif 3909  df-nul 4288  df-fr 5616

This theorem is referenced by: (None)