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Mirrors > Home > ILE Home > Th. List > sbbid | GIF version |
Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993.) |
Ref | Expression |
---|---|
sbbid.1 | ⊢ Ⅎ𝑥𝜑 |
sbbid.2 | ⊢ (𝜑 → (𝜓 ↔ 𝜒)) |
Ref | Expression |
---|---|
sbbid | ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sbbid.1 | . . 3 ⊢ Ⅎ𝑥𝜑 | |
2 | sbbid.2 | . . 3 ⊢ (𝜑 → (𝜓 ↔ 𝜒)) | |
3 | 1, 2 | alrimi 1456 | . 2 ⊢ (𝜑 → ∀𝑥(𝜓 ↔ 𝜒)) |
4 | spsbbi 1766 | . 2 ⊢ (∀𝑥(𝜓 ↔ 𝜒) → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒)) | |
5 | 3, 4 | syl 14 | 1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 103 ∀wal 1283 Ⅎwnf 1390 [wsb 1686 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-5 1377 ax-gen 1379 ax-ie1 1423 ax-ie2 1424 ax-4 1441 ax-ial 1468 |
This theorem depends on definitions: df-bi 115 df-nf 1391 df-sb 1687 |
This theorem is referenced by: bezoutlemmain 10520 |
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