Theorem sbbidv 2115

Description: Deduction substituting both sides of a biconditional, with 𝜑 and 𝑥 disjoint. See also sbbid 2284. (Contributed by Wolf Lammen, 6-May-2023.) (Proof shortened by Steven Nguyen, 6-Jul-2023.)

Hypothesis

Ref	Expression
sbbidv.1	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
sbbidv	⊢ (𝜑 → ([𝑡 / 𝑥]𝜓 ↔ [𝑡 / 𝑥]𝜒))

Distinct variable group:   𝜑,𝑥

Allowed substitution hints:   𝜑(𝑡)   𝜓(𝑥,𝑡)   𝜒(𝑥,𝑡)

Proof of Theorem sbbidv

Step	Hyp	Ref	Expression
1		sbbidv.1	. . 3 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
2	1	alrimiv 1950	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 ↔ 𝜒))
3		spsbbi 2109	. 2 ⊢ (∀𝑥(𝜓 ↔ 𝜒) → ([𝑡 / 𝑥]𝜓 ↔ [𝑡 / 𝑥]𝜒))
4	2, 3	syl 18	1 ⊢ (𝜑 → ([𝑡 / 𝑥]𝜓 ↔ [𝑡 / 𝑥]𝜒))

Syntax hints:   → wi 4   ↔ wb 209  ∀wal 1561  [wsb 2093

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933

This theorem depends on definitions:  df-bi 210  df-an 401  df-sb 2094

This theorem is referenced by:  sbco4lem  2138  sbco4  2139  sbcom2  2209  eqabdv  2898  wl-equsb3  38071  wl-clabtv  38101  2reu8i  47705  ichbidv  48057