Theorem sucel 6432

Description: Membership of a successor in another class. (Contributed by NM, 29-Jun-2004.)

Assertion

Ref	Expression
sucel	⊢ (suc 𝐴 ∈ 𝐵 ↔ ∃𝑥 ∈ 𝐵 ∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 = 𝐴)))

Distinct variable groups:   𝑥,𝑦,𝐴   𝑥,𝐵

Allowed substitution hint:   𝐵(𝑦)

Proof of Theorem sucel

Step	Hyp	Ref	Expression
1		risset 3238	. 2 ⊢ (suc 𝐴 ∈ 𝐵 ↔ ∃𝑥 ∈ 𝐵 𝑥 = suc 𝐴)
2		dfcleq 2754	. . . 4 ⊢ (𝑥 = suc 𝐴 ↔ ∀𝑦(𝑦 ∈ 𝑥 ↔ 𝑦 ∈ suc 𝐴))
3		vex 3455	. . . . . . 7 ⊢ 𝑦 ∈ V
4	3	elsuc 6428	. . . . . 6 ⊢ (𝑦 ∈ suc 𝐴 ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 = 𝐴))
5	4	bibi2i 340	. . . . 5 ⊢ ((𝑦 ∈ 𝑥 ↔ 𝑦 ∈ suc 𝐴) ↔ (𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 = 𝐴)))
6	5	albii 1852	. . . 4 ⊢ (∀𝑦(𝑦 ∈ 𝑥 ↔ 𝑦 ∈ suc 𝐴) ↔ ∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 = 𝐴)))
7	2, 6	bitri 278	. . 3 ⊢ (𝑥 = suc 𝐴 ↔ ∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 = 𝐴)))
8	7	rexbii 3110	. 2 ⊢ (∃𝑥 ∈ 𝐵 𝑥 = suc 𝐴 ↔ ∃𝑥 ∈ 𝐵 ∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 = 𝐴)))
9	1, 8	bitri 278	1 ⊢ (suc 𝐴 ∈ 𝐵 ↔ ∃𝑥 ∈ 𝐵 ∀𝑦(𝑦 ∈ 𝑥 ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 = 𝐴)))

This proof depends on syntax axioms:   ↔ wb 209   ∨ wo 861  ∀wal 1568   = wceq 1570   ∈ wcel 2145  ∃wrex 3087  suc csuc 6357

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-rex 3088  df-v 3453  df-un 3904  df-sn 4585  df-suc 6361

This theorem is used by:  axinf2  9625  zfinf2  9627