Theorem elsucg 6428

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 6363	. . . 4 ⊢ suc 𝐵 = (𝐵 ∪ {𝐵})
2	1	eleq2i 2852	. . 3 ⊢ (𝐴 ∈ suc 𝐵 ↔ 𝐴 ∈ (𝐵 ∪ {𝐵}))
3		elun 4100	. . 3 ⊢ (𝐴 ∈ (𝐵 ∪ {𝐵}) ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 ∈ {𝐵}))
4	2, 3	bitri 278	. 2 ⊢ (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 ∈ {𝐵}))
5		elsng 4598	. . 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 1570   ∈ wcel 2145   ∪ cun 3897  {csn 4584  suc csuc 6359

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 2732

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 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-un 3904  df-sn 4585  df-suc 6363

This theorem is used by:  elsuc  6430  elelsuc  6433  sucidg  6441  ordsssuc  6449  ordsucelsuc  7819  suc11reg  9601  nlt1pi  10918