Theorem elsuc 4405

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 4403	. 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 105   ∨ wo 708   = wceq 1353   ∈ wcel 2148  Vcvv 2737  suc csuc 4364

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2739  df-un 3133  df-sn 3598  df-suc 4370

This theorem is referenced by:  sucel  4409  suctr  4420  0elsucexmid  4563  tfrlemisucaccv  6323  tfr1onlemsucaccv  6339  tfrcllemsucaccv  6352