Theorem 2sbbid 2286

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 2285	. 2 ⊢ (𝜑 → ([𝑢 / 𝑦]𝜓 ↔ [𝑢 / 𝑦]𝜒))
5	1, 4	sbbid 2285	1 ⊢ (𝜑 → ([𝑡 / 𝑥][𝑢 / 𝑦]𝜓 ↔ [𝑡 / 𝑥][𝑢 / 𝑦]𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  Ⅎwnf 1816  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-12 2216

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by: (None)