Theorem dffr2ALT 5612

Description: Alternate proof of dffr2 5611, which avoids ax-8 2146 but requires ax-10 2177, ax-11 2193, ax-12 2212. (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 5603	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
2		rabeq0 4337	. . . . 5 ⊢ ({𝑧 ∈ 𝑥 ∣ 𝑧𝑅𝑦} = ∅ ↔ ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)
3	2	rexbii 3109	. . . 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 1851	. 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 1567   = wceq 1569   ≠ wne 2955  ∀wral 3076  ∃wrex 3086  {crab 3412   ⊆ wss 3895  ∅c0 4278   class class class wbr 5102   Fr wfr 5600

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-ral 3077  df-rex 3087  df-rab 3413  df-dif 4047  df-nul 4279  df-fr 5603

This theorem is used by: (None)