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Mirrors > Home > MPE Home > Th. List > Mathboxes > sbeqi | Structured version Visualization version GIF version |
Description: Equality deduction for substitution. (Contributed by Giovanni Mascellani, 10-Apr-2018.) |
Ref | Expression |
---|---|
sbeqi | ⊢ ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑 ↔ 𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | spsbbi 2069 | . 2 ⊢ (∀𝑧(𝜑 ↔ 𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓)) | |
2 | sbequ 2081 | . 2 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓)) | |
3 | 1, 2 | sylan9bbr 511 | 1 ⊢ ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑 ↔ 𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓)) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 207 ∧ wa 396 ∀wal 1526 = wceq 1528 [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 ax-6 1961 ax-7 2006 |
This theorem depends on definitions: df-bi 208 df-an 397 df-ex 1772 df-sb 2061 |
This theorem is referenced by: (None) |
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