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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 | pm3.4 316 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜑)) | |
2 | 19.8a 1482 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) | |
3 | df-sb 1646 | . . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))) | |
4 | 1, 2, 3 | sylanbrc 394 | . 2 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑) |
5 | 4 | ex 108 | 1 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 97 ∃wex 1381 [wsb 1645 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-4 1400 |
This theorem depends on definitions: df-bi 110 df-sb 1646 |
This theorem is referenced by: sbequ12 1654 sbequi 1720 sb6rf 1733 mo2n 1928 bj-bdfindes 10074 bj-findes 10106 |
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