Theorem bnj98 31265

Description: Technical lemma for bnj150 31274. 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 3343	. . . . . 6 ⊢ 𝑖 ∈ V
2	1	sucid 5965	. . . . 5 ⊢ 𝑖 ∈ suc 𝑖
3	2	n0ii 4065	. . . 4 ⊢ ¬ suc 𝑖 = ∅
4		df-suc 5890	. . . . . 6 ⊢ suc 𝑖 = (𝑖 ∪ {𝑖})
5		df-un 3720	. . . . . 6 ⊢ (𝑖 ∪ {𝑖}) = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
6	4, 5	eqtri 2782	. . . . 5 ⊢ suc 𝑖 = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
7		df1o2 7743	. . . . . . 7 ⊢ 1𝑜 = {∅}
8	6, 7	eleq12i 2832	. . . . . 6 ⊢ (suc 𝑖 ∈ 1𝑜 ↔ {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅})
9		elsni 4338	. . . . . 6 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅} → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
10	8, 9	sylbi 207	. . . . 5 ⊢ (suc 𝑖 ∈ 1𝑜 → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
11	6, 10	syl5eq 2806	. . . 4 ⊢ (suc 𝑖 ∈ 1𝑜 → suc 𝑖 = ∅)
12	3, 11	mto 188	. . 3 ⊢ ¬ suc 𝑖 ∈ 1𝑜
13	12	pm2.21i 116	. 2 ⊢ (suc 𝑖 ∈ 1𝑜 → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))
14	13	rgenw 3062	1 ⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1𝑜 → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Syntax hints:   → wi 4   ∨ wo 382   = wceq 1632   ∈ wcel 2139  {cab 2746  ∀wral 3050   ∪ cun 3713  ∅c0 4058  {csn 4321  ∪ ciun 4672  suc csuc 5886  ‘cfv 6049  ωcom 7231  1_𝑜c1o 7723   predc-bnj14 31084

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-nfc 2891  df-ral 3055  df-v 3342  df-dif 3718  df-un 3720  df-nul 4059  df-sn 4322  df-suc 5890  df-1o 7730

This theorem is referenced by:  bnj150  31274