Theorem sbbidv 2116

Description: Deduction substituting both sides of a biconditional, with 𝜑 and 𝑥 disjoint. See also sbbid 2285. (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 1960	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 ↔ 𝜒))
3		spsbbi 2110	. 2 ⊢ (∀𝑥(𝜓 ↔ 𝜒) → ([𝑡 / 𝑥]𝜓 ↔ [𝑡 / 𝑥]𝜒))
4	2, 3	syl 18	1 ⊢ (𝜑 → ([𝑡 / 𝑥]𝜓 ↔ [𝑡 / 𝑥]𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568  [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

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

This theorem is used by:  sbco4lem  2139  sbco4  2140  sbcom2  2210  eqabdv  2899  wl-equsb3  38252  wl-clabtv  38282  2reu8i  47891  ichbidv  48243