Theorem dffr2ALT 5628

Description: Alternate proof of dffr2 5627, which avoids ax-8 2148 but requires ax-10 2179, ax-11 2195, ax-12 2216. (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 5619	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
2		rabeq0 4348	. . . . 5 ⊢ ({𝑧 ∈ 𝑥 ∣ 𝑧𝑅𝑦} = ∅ ↔ ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)
3	2	rexbii 3115	. . . 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 1852	. 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} = ∅))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570   ≠ wne 2961  ∀wral 3082  ∃wrex 3092  {crab 3419   ⊆ wss 3908  ∅c0 4289   class class class wbr 5114   Fr wfr 5616

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2745  df-cleq 2758  df-ral 3083  df-rex 3093  df-rab 3420  df-dif 3911  df-nul 4290  df-fr 5619

This theorem is used by: (None)