Theorem elsuc 4143

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 4141	. 2 ⊢ (𝐴 ∈ V → (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵)))
3	1, 2	ax-mp 7	1 ⊢ (𝐴 ∈ suc 𝐵 ↔ (𝐴 ∈ 𝐵 ∨ 𝐴 = 𝐵))

Syntax hints:   ↔ wb 98   ∨ wo 629   = wceq 1243   ∈ wcel 1393  Vcvv 2557  suc csuc 4102

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-un 2922  df-sn 3381  df-suc 4108

This theorem is referenced by:  sucel  4147  suctr  4158  0elsucexmid  4289  tfrlemisucaccv  5939