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Mirrors > Home > ILE Home > Th. List > sbequ1 | GIF version |
Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.) |
Ref | Expression |
---|---|
sbequ1 | ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | pm3.4 333 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜑)) | |
2 | 19.8a 1590 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) | |
3 | df-sb 1763 | . . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))) | |
4 | 1, 2, 3 | sylanbrc 417 | . 2 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑) |
5 | 4 | ex 115 | 1 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 104 ∃wex 1492 [wsb 1762 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-gen 1449 ax-ie1 1493 ax-ie2 1494 ax-4 1510 |
This theorem depends on definitions: df-bi 117 df-sb 1763 |
This theorem is referenced by: sbequ12 1771 sbequi 1839 sb6rf 1853 mo2n 2054 bj-bdfindes 14583 bj-findes 14615 |
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