Theorem bnj98 34881

Description: Technical lemma for bnj150 34890. 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 3484	. . . . . 6 ⊢ 𝑖 ∈ V
2	1	sucid 6466	. . . . 5 ⊢ 𝑖 ∈ suc 𝑖
3	2	n0ii 4343	. . . 4 ⊢ ¬ suc 𝑖 = ∅
4		df-suc 6390	. . . . . 6 ⊢ suc 𝑖 = (𝑖 ∪ {𝑖})
5		df-un 3956	. . . . . 6 ⊢ (𝑖 ∪ {𝑖}) = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
6	4, 5	eqtri 2765	. . . . 5 ⊢ suc 𝑖 = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
7		df1o2 8513	. . . . . . 7 ⊢ 1o = {∅}
8	6, 7	eleq12i 2834	. . . . . 6 ⊢ (suc 𝑖 ∈ 1o ↔ {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅})
9		elsni 4643	. . . . . 6 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅} → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
10	8, 9	sylbi 217	. . . . 5 ⊢ (suc 𝑖 ∈ 1o → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
11	6, 10	eqtrid 2789	. . . 4 ⊢ (suc 𝑖 ∈ 1o → suc 𝑖 = ∅)
12	3, 11	mto 197	. . 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 3065	1 ⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1o → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Syntax hints:   → wi 4   ∨ wo 848   = wceq 1540   ∈ wcel 2108  {cab 2714  ∀wral 3061   ∪ cun 3949  ∅c0 4333  {csn 4626  ∪ ciun 4991  suc csuc 6386  ‘cfv 6561  ωcom 7887  1_oc1o 8499   predc-bnj14 34702

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3062  df-v 3482  df-dif 3954  df-un 3956  df-nul 4334  df-sn 4627  df-suc 6390  df-1o 8506

This theorem is referenced by:  bnj150  34890