Theorem bnj98 30191

Description: Technical lemma for bnj150 30200. 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 3176	. . . . . 6 ⊢ 𝑖 ∈ V
2	1	sucid 5721	. . . . 5 ⊢ 𝑖 ∈ suc 𝑖
3	2	n0ii 3881	. . . 4 ⊢ ¬ suc 𝑖 = ∅
4		df-suc 5646	. . . . . 6 ⊢ suc 𝑖 = (𝑖 ∪ {𝑖})
5		df-un 3545	. . . . . 6 ⊢ (𝑖 ∪ {𝑖}) = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
6	4, 5	eqtri 2632	. . . . 5 ⊢ suc 𝑖 = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
7		df1o2 7459	. . . . . . 7 ⊢ 1𝑜 = {∅}
8	6, 7	eleq12i 2681	. . . . . 6 ⊢ (suc 𝑖 ∈ 1𝑜 ↔ {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅})
9		elsni 4142	. . . . . 6 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅} → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
10	8, 9	sylbi 206	. . . . 5 ⊢ (suc 𝑖 ∈ 1𝑜 → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
11	6, 10	syl5eq 2656	. . . 4 ⊢ (suc 𝑖 ∈ 1𝑜 → suc 𝑖 = ∅)
12	3, 11	mto 187	. . 3 ⊢ ¬ suc 𝑖 ∈ 1𝑜
13	12	pm2.21i 115	. 2 ⊢ (suc 𝑖 ∈ 1𝑜 → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))
14	13	rgenw 2908	1 ⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1𝑜 → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Syntax hints:   → wi 4   ∨ wo 382   = wceq 1475   ∈ wcel 1977  {cab 2596  ∀wral 2896   ∪ cun 3538  ∅c0 3874  {csn 4125  ∪ ciun 4455  suc csuc 5642  ‘cfv 5804  ωcom 6957  1_𝑜c1o 7440   predc-bnj14 30007

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ral 2901  df-v 3175  df-dif 3543  df-un 3545  df-nul 3875  df-sn 4126  df-suc 5646  df-1o 7447

This theorem is referenced by:  bnj150  30200