Theorem sbcbid 3055

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

Hypotheses

Ref	Expression
sbcbid.1	⊢ Ⅎ𝑥𝜑
sbcbid.2	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

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

Proof of Theorem sbcbid

Step	Hyp	Ref	Expression
1		sbcbid.1	. . . 4 ⊢ Ⅎ𝑥𝜑
2		sbcbid.2	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
3	1, 2	abbid 2321	. . 3 ⊢ (𝜑 → {𝑥 ∣ 𝜓} = {𝑥 ∣ 𝜒})
4	3	eleq2d 2274	. 2 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝐴 ∈ {𝑥 ∣ 𝜒}))
5		df-sbc 2998	. 2 ⊢ ([𝐴 / 𝑥]𝜓 ↔ 𝐴 ∈ {𝑥 ∣ 𝜓})
6		df-sbc 2998	. 2 ⊢ ([𝐴 / 𝑥]𝜒 ↔ 𝐴 ∈ {𝑥 ∣ 𝜒})
7	4, 5, 6	3bitr4g 223	1 ⊢ (𝜑 → ([𝐴 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜒))

Syntax hints:   → wi 4   ↔ wb 105  Ⅎwnf 1482   ∈ wcel 2175  {cab 2190  [wsbc 2997

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 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-11 1528  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-tru 1375  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-sbc 2998

This theorem is referenced by:  sbcbidv  3056  csbeq2d  3117  bezoutlemstep  12289