Theorem elsuc 6453

Description: Membership in a successor. Exercise 5 of [TakeutiZaring] p. 17. (Contributed by NM, 15-Sep-2003.)

Hypothesis

Ref	Expression
elsuc.1	⊢ 𝐴 ∈ V

Assertion

Ref	Expression
elsuc	⊢ (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵))

Proof of Theorem elsuc

Step	Hyp	Ref	Expression
1		elsuc.1	. 2 ⊢ 𝐴 ∈ V
2		elsucg 6451	. 2 ⊢ (𝐴 ∈ V → (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵)))
3	1, 2	ax-mp 5	1 ⊢ (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵))

This proof depends on syntax axioms:   ↔ wb 209   ∨ wo 861   = wceq 1570   ∈ wcel 2147  Vcvv 3454  suc csuc 6382

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-un 4154  df-sn 4589  df-suc 6386

This theorem is used by:  sucel  6457  limsssuc  7885  omsmolem  8694  cantnfle  9699  infxpenlem  10119  inatsk  10890  nolesgn2ores  28059  nogesgn1ores  28061  untsucf  36653  dfon2lem7  36729  rdgssun  38493  omssaxinf2  46173