Theorem bnj98 35264

Description: Technical lemma for bnj150 35273. 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 3458	. . . . . 6 ⊢ 𝑖 ∈ V
2	1	sucid 6445	. . . . 5 ⊢ 𝑖 ∈ suc 𝑖
3	2	n0ii 4295	. . . 4 ⊢ ¬ suc 𝑖 = ∅
4		df-suc 6366	. . . . . 6 ⊢ suc 𝑖 = (𝑖 ∪ {𝑖})
5		df-un 3909	. . . . . 6 ⊢ (𝑖 ∪ {𝑖}) = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
6	4, 5	eqtri 2785	. . . . 5 ⊢ suc 𝑖 = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
7		df1o2 8458	. . . . . . 7 ⊢ 1o = {∅}
8	6, 7	eleq12i 2855	. . . . . 6 ⊢ (suc 𝑖 ∈ 1o ↔ {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅})
9		elsni 4605	. . . . . 6 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅} → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
10	8, 9	sylbi 220	. . . . 5 ⊢ (suc 𝑖 ∈ 1o → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
11	6, 10	eqtrid 2809	. . . 4 ⊢ (suc 𝑖 ∈ 1o → suc 𝑖 = ∅)
12	3, 11	mto 200	. . 3 ⊢ ¬ suc 𝑖 ∈ 1o
13	12	pm2.21i 120	. 2 ⊢ (suc 𝑖 ∈ 1o → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))
14	13	rgenw 3082	1 ⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1o → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

This proof depends on syntax axioms:   → wi 4   ∨ wo 860   = wceq 1569   ∈ wcel 2142  {cab 2740  ∀wral 3078   ∪ cun 3902  ∅c0 4285  {csn 4588  ∪ ciun 4955  suc csuc 6362  ‘cfv 6536  ωcom 7860  1_oc1o 8444   predc-bnj14 35086

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-v 3456  df-dif 3907  df-un 3909  df-nul 4286  df-sn 4589  df-suc 6366  df-1o 8451

This theorem is used by:  bnj150  35273