Theorem sbcbidv 3110

Description: Formula-building deduction for class substitution. (Contributed by NM, 29-Dec-2014.)

Hypothesis

Ref	Expression
sbcbidv.1	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
sbcbidv	⊢ (𝜑 → ([𝐴 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜒))

Distinct variable group:   𝜑,𝑥

Allowed substitution hints:   𝜓(𝑥)   𝜒(𝑥)   𝐴(𝑥)

Proof of Theorem sbcbidv

Step	Hyp	Ref	Expression
1		nfv 1581	. 2 ⊢ Ⅎ𝑥𝜑
2		sbcbidv.1	. 2 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
3	1, 2	sbcbid 3109	1 ⊢ (𝜑 → ([𝐴 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜒))

Syntax hints:   → wi 4   ↔ wb 105  [wsbc 3051

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-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-11 1559  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-sbc 3052

This theorem is referenced by:  sbcbii  3111  csbcomg  3170  opelopabsb  4397  opelopabgf  4407  opelopabf  4412  sbcfng  5526  sbcfg  5527  uchoice  6361  f1od2  6461  wrd2ind  11473  islmod  14600