Theorem elsuc 6390

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 6388	. 2 ⊢ (𝐴 ∈ V → (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵)))
3	1, 2	ax-mp 5	1 ⊢ (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵))

Syntax hints:   ↔ wb 206   ∨ wo 848   = wceq 1542   ∈ wcel 2114  Vcvv 3441  suc csuc 6320

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-v 3443  df-un 3907  df-sn 4582  df-suc 6324

This theorem is referenced by:  sucel  6394  limsssuc  7794  omsmolem  8587  cantnfle  9584  infxpenlem  9927  inatsk  10693  nolesgn2ores  27644  nogesgn1ores  27646  untsucf  35906  dfon2lem7  35983  rdgssun  37585  omssaxinf2  45296