Theorem sbbidh 1859

Description: Deduction substituting both sides of a biconditional. New proofs should use sbbid 1860 instead. (Contributed by NM, 5-Aug-1993.) (New usage is discouraged.)

Hypotheses

Ref	Expression
sbbidh.1	⊢ (𝜑 → ∀𝑥𝜑)
sbbidh.2	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
sbbidh	⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Proof of Theorem sbbidh

Step	Hyp	Ref	Expression
1		sbbidh.1	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
2		sbbidh.2	. . 3 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
3	1, 2	alrimih 1483	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 ↔ 𝜒))
4		spsbbi 1858	. 2 ⊢ (∀𝑥(𝜓 ↔ 𝜒) → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))
5	3, 4	syl 14	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Syntax hints:   → wi 4   ↔ wb 105  ∀wal 1362  [wsb 1776

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 1461  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-4 1524  ax-ial 1548

This theorem depends on definitions:  df-bi 117  df-sb 1777

This theorem is referenced by:  sbcomxyyz  1991