Theorem bnj98 31751

Description: Technical lemma for bnj150 31760. 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 3443	. . . . . 6 ⊢ 𝑖 ∈ V
2	1	sucid 6152	. . . . 5 ⊢ 𝑖 ∈ suc 𝑖
3	2	n0ii 4228	. . . 4 ⊢ ¬ suc 𝑖 = ∅
4		df-suc 6079	. . . . . 6 ⊢ suc 𝑖 = (𝑖 ∪ {𝑖})
5		df-un 3870	. . . . . 6 ⊢ (𝑖 ∪ {𝑖}) = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
6	4, 5	eqtri 2821	. . . . 5 ⊢ suc 𝑖 = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
7		df1o2 7974	. . . . . . 7 ⊢ 1o = {∅}
8	6, 7	eleq12i 2877	. . . . . 6 ⊢ (suc 𝑖 ∈ 1o ↔ {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅})
9		elsni 4495	. . . . . 6 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅} → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
10	8, 9	sylbi 218	. . . . 5 ⊢ (suc 𝑖 ∈ 1o → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
11	6, 10	syl5eq 2845	. . . 4 ⊢ (suc 𝑖 ∈ 1o → suc 𝑖 = ∅)
12	3, 11	mto 198	. . 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 3119	1 ⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1o → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Syntax hints:   → wi 4   ∨ wo 842   = wceq 1525   ∈ wcel 2083  {cab 2777  ∀wral 3107   ∪ cun 3863  ∅c0 4217  {csn 4478  ∪ ciun 4831  suc csuc 6075  ‘cfv 6232  ωcom 7443  1_oc1o 7953   predc-bnj14 31571

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1781  ax-4 1795  ax-5 1892  ax-6 1951  ax-7 1996  ax-8 2085  ax-9 2093  ax-10 2114  ax-11 2128  ax-12 2143  ax-ext 2771

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-tru 1528  df-ex 1766  df-nf 1770  df-sb 2045  df-clab 2778  df-cleq 2790  df-clel 2865  df-nfc 2937  df-ral 3112  df-v 3442  df-dif 3868  df-un 3870  df-nul 4218  df-sn 4479  df-suc 6079  df-1o 7960

This theorem is referenced by:  bnj150  31760