Theorem elsuc 6434

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 6432	. 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 2145  Vcvv 3453  suc csuc 6363

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 2734

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 2741  df-cleq 2754  df-clel 2837  df-v 3455  df-un 3907  df-sn 4588  df-suc 6367

This theorem is used by:  sucel  6438  limsssuc  7849  omsmolem  8648  cantnfle  9653  infxpenlem  10019  inatsk  10790  nolesgn2ores  27906  nogesgn1ores  27908  untsucf  36276  dfon2lem7  36353  rdgssun  38119  omssaxinf2  45798