Theorem sbcbidv 3101

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 1577	. 2 ⊢ Ⅎ𝑥𝜑
2		sbcbidv.1	. 2 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
3	1, 2	sbcbid 3100	1 ⊢ (𝜑 → ([𝐴 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜒))

Syntax hints:   → wi 4   ↔ wb 105  [wsbc 3042

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 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-11 1555  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-sbc 3043

This theorem is referenced by:  sbcbii  3102  csbcomg  3161  opelopabsb  4378  opelopabgf  4388  opelopabf  4393  sbcfng  5506  sbcfg  5507  uchoice  6331  f1od2  6431  wrd2ind  11415  islmod  14439