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Mirrors > Home > MPE Home > Th. List > Mathboxes > bj-dfsb2 | Structured version Visualization version GIF version |
Description: Alternate (dual) definition of substitution df-sb 2070 not using dummy variables. (Contributed by BJ, 19-Mar-2021.) |
Ref | Expression |
---|---|
bj-dfsb2 | ⊢ ([𝑦 / 𝑥]𝜑 ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | dfsb1 2510 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))) | |
2 | bj-sbsb 34162 | . 2 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑))) | |
3 | 1, 2 | bitri 277 | 1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 208 ∧ wa 398 ∨ wo 843 ∀wal 1535 ∃wex 1780 [wsb 2069 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-10 2145 ax-12 2177 ax-13 2390 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-ex 1781 df-nf 1785 df-sb 2070 |
This theorem is referenced by: (None) |
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