Theorem sbcbi2 3001

Description: Substituting into equivalent wff's gives equivalent results. (Contributed by Giovanni Mascellani, 9-Apr-2018.)

Assertion

Ref	Expression
sbcbi2	⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))

Proof of Theorem sbcbi2

Step	Hyp	Ref	Expression
1		abbi 2280	. . 3 ⊢ (∀𝑥(𝜑 ↔ 𝜓) ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓})
2		eleq2 2230	. . 3 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜓}))
3	1, 2	sylbi 120	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜓}))
4		df-sbc 2952	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
5		df-sbc 2952	. 2 ⊢ ([𝐴 / 𝑥]𝜓 ↔ 𝐴 ∈ {𝑥 ∣ 𝜓})
6	3, 4, 5	3bitr4g 222	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))

Syntax hints:   → wi 4   ↔ wb 104  ∀wal 1341   = wceq 1343   ∈ wcel 2136  {cab 2151  [wsbc 2951

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-11 1494  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-sbc 2952

This theorem is referenced by:  csbeq2  3069