Theorem elsucg 6439

Description: Membership in a successor. Exercise 5 of [TakeutiZaring] p. 17. (Contributed by NM, 15-Sep-1995.)

Assertion

Ref	Expression
elsucg	⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵)))

Proof of Theorem elsucg

Step	Hyp	Ref	Expression
1		df-suc 6374	. . . 4 ⊢ suc 𝐵 = (𝐵 ∪ {𝐵})
2	1	eleq2i 2855	. . 3 ⊢ (𝐴 ∈ suc 𝐵 ↔ 𝐴 ∈ (𝐵 ∪ {𝐵}))
3		elun 4154	. . 3 ⊢ (𝐴 ∈ (𝐵 ∪ {𝐵}) ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 ∈ {𝐵}))
4	2, 3	bitri 278	. 2 ⊢ (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 ∈ {𝐵}))
5		elsng 4597	. . 3 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝐵} ↔ 𝐴 = 𝐵))
6	5	orbi2d 929	. 2 ⊢ (𝐴 ∈ 𝑉 → ((𝐴 ∈ 𝐵 ∨ 𝐴 ∈ {𝐵}) ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵)))
7	4, 6	bitrid 286	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∨ wo 861   = wceq 1569   ∈ wcel 2146   ∪ cun 4151  {csn 4587  suc csuc 6370

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3455  df-un 4153  df-sn 4588  df-suc 6374

This theorem is used by:  elsuc  6441  elelsuc  6444  sucidg  6452  ordsssuc  6460  ordsucelsuc  7840  suc11reg  9630  nlt1pi  11001