Theorem 2sbbid 2250

Description: Deduction doubly substituting both sides of a biconditional. (Contributed by AV, 30-Jul-2023.)

Hypotheses

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

Assertion

Ref	Expression
2sbbid	⊢ (𝜑 → ([𝑡 / 𝑥][𝑢 / 𝑦]𝜓 ↔ [𝑡 / 𝑥][𝑢 / 𝑦]𝜒))

Proof of Theorem 2sbbid

Step	Hyp	Ref	Expression
1		sbbid.1	. 2 ⊢ Ⅎ𝑥𝜑
2		2sbbid.1	. . 3 ⊢ Ⅎ𝑦𝜑
3		sbbid.2	. . 3 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
4	2, 3	sbbid 2249	. 2 ⊢ (𝜑 → ([𝑢 / 𝑦]𝜓 ↔ [𝑢 / 𝑦]𝜒))
5	1, 4	sbbid 2249	1 ⊢ (𝜑 → ([𝑡 / 𝑥][𝑢 / 𝑦]𝜓 ↔ [𝑡 / 𝑥][𝑢 / 𝑦]𝜒))

Syntax hints:   → wi 4   ↔ wb 206  Ⅎwnf 1784  [wsb 2067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-12 2180

This theorem depends on definitions:  df-bi 207  df-ex 1781  df-nf 1785  df-sb 2068

This theorem is referenced by: (None)