Theorem bnj98 31260

Description: Technical lemma for bnj150 31269. 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 3394	. . . . . 6 ⊢ 𝑖 ∈ V
2	1	sucid 6017	. . . . 5 ⊢ 𝑖 ∈ suc 𝑖
3	2	n0ii 4124	. . . 4 ⊢ ¬ suc 𝑖 = ∅
4		df-suc 5942	. . . . . 6 ⊢ suc 𝑖 = (𝑖 ∪ {𝑖})
5		df-un 3774	. . . . . 6 ⊢ (𝑖 ∪ {𝑖}) = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
6	4, 5	eqtri 2828	. . . . 5 ⊢ suc 𝑖 = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
7		df1o2 7809	. . . . . . 7 ⊢ 1𝑜 = {∅}
8	6, 7	eleq12i 2878	. . . . . 6 ⊢ (suc 𝑖 ∈ 1𝑜 ↔ {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅})
9		elsni 4387	. . . . . 6 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅} → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
10	8, 9	sylbi 208	. . . . 5 ⊢ (suc 𝑖 ∈ 1𝑜 → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
11	6, 10	syl5eq 2852	. . . 4 ⊢ (suc 𝑖 ∈ 1𝑜 → suc 𝑖 = ∅)
12	3, 11	mto 188	. . 3 ⊢ ¬ suc 𝑖 ∈ 1𝑜
13	12	pm2.21i 117	. 2 ⊢ (suc 𝑖 ∈ 1𝑜 → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))
14	13	rgenw 3112	1 ⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1𝑜 → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Syntax hints:   → wi 4   ∨ wo 865   = wceq 1637   ∈ wcel 2156  {cab 2792  ∀wral 3096   ∪ cun 3767  ∅c0 4116  {csn 4370  ∪ ciun 4712  suc csuc 5938  ‘cfv 6101  ωcom 7295  1_𝑜c1o 7789   predc-bnj14 31079

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2068  ax-7 2104  ax-9 2165  ax-10 2185  ax-11 2201  ax-12 2214  ax-13 2420  ax-ext 2784

This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2061  df-clab 2793  df-cleq 2799  df-clel 2802  df-nfc 2937  df-ral 3101  df-v 3393  df-dif 3772  df-un 3774  df-nul 4117  df-sn 4371  df-suc 5942  df-1o 7796

This theorem is referenced by:  bnj150  31269