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Theorem sbbidv 2075
Description: Deduction substituting both sides of a biconditional, with 𝜑 and 𝑥 disjoint. See also sbbid 2236. (Contributed by Wolf Lammen, 6-May-2023.) (Proof shortened by Steven Nguyen, 6-Jul-2023.)
Hypothesis
Ref Expression
sbbidv.1 (𝜑 → (𝜓𝜒))
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
StepHypRef Expression
1 sbbidv.1 . . 3 (𝜑 → (𝜓𝜒))
21alrimiv 1919 . 2 (𝜑 → ∀𝑥(𝜓𝜒))
3 spsbbi 2069 . 2 (∀𝑥(𝜓𝜒) → ([𝑡 / 𝑥]𝜓 ↔ [𝑡 / 𝑥]𝜒))
42, 3syl 17 1 (𝜑 → ([𝑡 / 𝑥]𝜓 ↔ [𝑡 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207  wal 1526  [wsb 2060
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902
This theorem depends on definitions:  df-bi 208  df-sb 2061
This theorem is referenced by:  sbcom2  2158  abbi2dv  2947  wl-equsb3  34673  wl-clabtv  34710  2reu8i  43189
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