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Mirrors > Home > MPE Home > Th. List > Mathboxes > sbequbidv | Structured version Visualization version GIF version |
Description: Deduction substituting both sides of a biconditional. (Contributed by GG, 1-Sep-2025.) |
Ref | Expression |
---|---|
sbequbidv.1 | ⊢ (𝜑 → 𝑢 = 𝑣) |
sbequbidv.2 | ⊢ (𝜑 → (𝜓 ↔ 𝜒)) |
Ref | Expression |
---|---|
sbequbidv | ⊢ (𝜑 → ([𝑢 / 𝑥]𝜓 ↔ [𝑣 / 𝑥]𝜒)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sbequbidv.1 | . . . . 5 ⊢ (𝜑 → 𝑢 = 𝑣) | |
2 | equequ2 2021 | . . . . 5 ⊢ (𝑢 = 𝑣 → (𝑡 = 𝑢 ↔ 𝑡 = 𝑣)) | |
3 | 1, 2 | syl 17 | . . . 4 ⊢ (𝜑 → (𝑡 = 𝑢 ↔ 𝑡 = 𝑣)) |
4 | sbequbidv.2 | . . . . . 6 ⊢ (𝜑 → (𝜓 ↔ 𝜒)) | |
5 | 4 | imbi2d 340 | . . . . 5 ⊢ (𝜑 → ((𝑥 = 𝑡 → 𝜓) ↔ (𝑥 = 𝑡 → 𝜒))) |
6 | 5 | albidv 1916 | . . . 4 ⊢ (𝜑 → (∀𝑥(𝑥 = 𝑡 → 𝜓) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜒))) |
7 | 3, 6 | imbi12d 344 | . . 3 ⊢ (𝜑 → ((𝑡 = 𝑢 → ∀𝑥(𝑥 = 𝑡 → 𝜓)) ↔ (𝑡 = 𝑣 → ∀𝑥(𝑥 = 𝑡 → 𝜒)))) |
8 | 7 | albidv 1916 | . 2 ⊢ (𝜑 → (∀𝑡(𝑡 = 𝑢 → ∀𝑥(𝑥 = 𝑡 → 𝜓)) ↔ ∀𝑡(𝑡 = 𝑣 → ∀𝑥(𝑥 = 𝑡 → 𝜒)))) |
9 | df-sb 2061 | . 2 ⊢ ([𝑢 / 𝑥]𝜓 ↔ ∀𝑡(𝑡 = 𝑢 → ∀𝑥(𝑥 = 𝑡 → 𝜓))) | |
10 | df-sb 2061 | . 2 ⊢ ([𝑣 / 𝑥]𝜒 ↔ ∀𝑡(𝑡 = 𝑣 → ∀𝑥(𝑥 = 𝑡 → 𝜒))) | |
11 | 8, 9, 10 | 3bitr4g 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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