Theorem bnj98 32826

Description: Technical lemma for bnj150 32835. 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 3434	. . . . . 6 ⊢ 𝑖 ∈ V
2	1	sucid 6342	. . . . 5 ⊢ 𝑖 ∈ suc 𝑖
3	2	n0ii 4275	. . . 4 ⊢ ¬ suc 𝑖 = ∅
4		df-suc 6269	. . . . . 6 ⊢ suc 𝑖 = (𝑖 ∪ {𝑖})
5		df-un 3896	. . . . . 6 ⊢ (𝑖 ∪ {𝑖}) = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
6	4, 5	eqtri 2767	. . . . 5 ⊢ suc 𝑖 = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
7		df1o2 8293	. . . . . . 7 ⊢ 1o = {∅}
8	6, 7	eleq12i 2832	. . . . . 6 ⊢ (suc 𝑖 ∈ 1o ↔ {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅})
9		elsni 4583	. . . . . 6 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅} → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
10	8, 9	sylbi 216	. . . . 5 ⊢ (suc 𝑖 ∈ 1o → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
11	6, 10	eqtrid 2791	. . . 4 ⊢ (suc 𝑖 ∈ 1o → suc 𝑖 = ∅)
12	3, 11	mto 196	. . 3 ⊢ ¬ suc 𝑖 ∈ 1o
13	12	pm2.21i 119	. 2 ⊢ (suc 𝑖 ∈ 1o → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))
14	13	rgenw 3077	1 ⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1o → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Syntax hints:   → wi 4   ∨ wo 843   = wceq 1541   ∈ wcel 2109  {cab 2716  ∀wral 3065   ∪ cun 3889  ∅c0 4261  {csn 4566  ∪ ciun 4929  suc csuc 6265  ‘cfv 6430  ωcom 7700  1_oc1o 8274   predc-bnj14 32646

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-ext 2710

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1544  df-fal 1554  df-ex 1786  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-ral 3070  df-v 3432  df-dif 3894  df-un 3896  df-nul 4262  df-sn 4567  df-suc 6269  df-1o 8281

This theorem is referenced by:  bnj150  32835