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Theorem sbequbidv 36157
Description: Deduction substituting both sides of a biconditional. (Contributed by GG, 1-Sep-2025.)
Hypotheses
Ref Expression
sbequbidv.1 (𝜑𝑢 = 𝑣)
sbequbidv.2 (𝜑 → (𝜓𝜒))
Assertion
Ref Expression
sbequbidv (𝜑 → ([𝑢 / 𝑥]𝜓 ↔ [𝑣 / 𝑥]𝜒))
Distinct variable group:   𝜑,𝑥
Allowed substitution hints:   𝜑(𝑣,𝑢)   𝜓(𝑥,𝑣,𝑢)   𝜒(𝑥,𝑣,𝑢)

Proof of Theorem sbequbidv
Dummy variable 𝑡 is distinct from all other variables.
StepHypRef Expression
1 sbequbidv.1 . . . . 5 (𝜑𝑢 = 𝑣)
2 equequ2 2021 . . . . 5 (𝑢 = 𝑣 → (𝑡 = 𝑢𝑡 = 𝑣))
31, 2syl 17 . . . 4 (𝜑 → (𝑡 = 𝑢𝑡 = 𝑣))
4 sbequbidv.2 . . . . . 6 (𝜑 → (𝜓𝜒))
54imbi2d 340 . . . . 5 (𝜑 → ((𝑥 = 𝑡𝜓) ↔ (𝑥 = 𝑡𝜒)))
65albidv 1916 . . . 4 (𝜑 → (∀𝑥(𝑥 = 𝑡𝜓) ↔ ∀𝑥(𝑥 = 𝑡𝜒)))
73, 6imbi12d 344 . . 3 (𝜑 → ((𝑡 = 𝑢 → ∀𝑥(𝑥 = 𝑡𝜓)) ↔ (𝑡 = 𝑣 → ∀𝑥(𝑥 = 𝑡𝜒))))
87albidv 1916 . 2 (𝜑 → (∀𝑡(𝑡 = 𝑢 → ∀𝑥(𝑥 = 𝑡𝜓)) ↔ ∀𝑡(𝑡 = 𝑣 → ∀𝑥(𝑥 = 𝑡𝜒))))
9 df-sb 2061 . 2 ([𝑢 / 𝑥]𝜓 ↔ ∀𝑡(𝑡 = 𝑢 → ∀𝑥(𝑥 = 𝑡𝜓)))
10 df-sb 2061 . 2 ([𝑣 / 𝑥]𝜒 ↔ ∀𝑡(𝑡 = 𝑣 → ∀𝑥(𝑥 = 𝑡𝜒)))
118, 9, 103bitr4g 314 1 (𝜑 → ([𝑢 / 𝑥]𝜓 ↔ [𝑣 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wal 1533  [wsb 2060
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1963  ax-7 2003
This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1775  df-sb 2061
This theorem is referenced by: (None)
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