Theorem sbc7 3770

Description: An equivalence for class substitution in the spirit of df-clab 2739. Note that 𝑥 and 𝐴 don't have to be distinct. (Contributed by NM, 18-Nov-2008.) (Revised by Mario Carneiro, 13-Oct-2016.)

Assertion

Ref	Expression
sbc7	⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑦(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))

Distinct variable groups:   𝑦,𝐴   𝜑,𝑦   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥)   𝐴(𝑥)

Proof of Theorem sbc7

Step	Hyp	Ref	Expression
1		sbccow 3761	. 2 ⊢ ([𝐴 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑)
2		sbc5 3766	. 2 ⊢ ([𝐴 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∃𝑦(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
3	1, 2	bitr3i 280	1 ⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑦(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1569  ∃wex 1811  [wsbc 3738

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-sbc 3739

This theorem is used by:  bj-df-sb  37643