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Mirrors > Home > ILE Home > Th. List > sb6x | GIF version |
Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 12-Aug-2011.) |
Ref | Expression |
---|---|
sb6x.1 | ⊢ (φ → ∀xφ) |
Ref | Expression |
---|---|
sb6x | ⊢ ([y / x]φ ↔ ∀x(x = y → φ)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sb6x.1 | . . 3 ⊢ (φ → ∀xφ) | |
2 | 1 | sbh 1656 | . 2 ⊢ ([y / x]φ ↔ φ) |
3 | biidd 161 | . . 3 ⊢ (x = y → (φ ↔ φ)) | |
4 | 1, 3 | equsalh 1611 | . 2 ⊢ (∀x(x = y → φ) ↔ φ) |
5 | 2, 4 | bitr4i 176 | 1 ⊢ ([y / x]φ ↔ ∀x(x = y → φ)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 98 ∀wal 1240 [wsb 1642 |
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-5 1333 ax-gen 1335 ax-ie1 1379 ax-ie2 1380 ax-4 1397 ax-i9 1420 ax-ial 1424 |
This theorem depends on definitions: df-bi 110 df-sb 1643 |
This theorem is referenced by: (None) |
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