Theorem bnj98 35225

Description: Technical lemma for bnj150 35234. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)

Assertion

Ref	Expression
bnj98	⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1o → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Proof of Theorem bnj98

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 3466	. . . . . 6 ⊢ 𝑖 ∈ V
2	1	sucid 6449	. . . . 5 ⊢ 𝑖 ∈ suc 𝑖
3	2	n0ii 4304	. . . 4 ⊢ ¬ suc 𝑖 = ∅
4		df-suc 6370	. . . . . 6 ⊢ suc 𝑖 = (𝑖 ∪ {𝑖})
5		df-un 3918	. . . . . 6 ⊢ (𝑖 ∪ {𝑖}) = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
6	4, 5	eqtri 2793	. . . . 5 ⊢ suc 𝑖 = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
7		df1o2 8463	. . . . . . 7 ⊢ 1o = {∅}
8	6, 7	eleq12i 2863	. . . . . 6 ⊢ (suc 𝑖 ∈ 1o ↔ {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅})
9		elsni 4611	. . . . . 6 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅} → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
10	8, 9	sylbi 220	. . . . 5 ⊢ (suc 𝑖 ∈ 1o → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
11	6, 10	eqtrid 2817	. . . 4 ⊢ (suc 𝑖 ∈ 1o → suc 𝑖 = ∅)
12	3, 11	mto 200	. . 3 ⊢ ¬ suc 𝑖 ∈ 1o
13	12	pm2.21i 120	. 2 ⊢ (suc 𝑖 ∈ 1o → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))
14	13	rgenw 3090	1 ⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1o → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Syntax hints:   → wi 4   ∨ wo 860   = wceq 1568   ∈ wcel 2150  {cab 2748  ∀wral 3086   ∪ cun 3911  ∅c0 4294  {csn 4594  ∪ ciun 4961  suc csuc 6366  ‘cfv 6540  ωcom 7865  1_oc1o 8449   predc-bnj14 35047

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2152  ax-9 2160  ax-ext 2742

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1571  df-fal 1581  df-ex 1808  df-sb 2099  df-clab 2749  df-cleq 2762  df-clel 2845  df-ral 3087  df-v 3464  df-dif 3916  df-un 3918  df-nul 4295  df-sn 4595  df-suc 6370  df-1o 8456

This theorem is referenced by:  bnj150  35234