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Theorem sb6x 2029
 Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)
Hypothesis
Ref Expression
sb6x.1 xφ
Assertion
Ref Expression
sb6x ([y / x]φx(x = yφ))

Proof of Theorem sb6x
StepHypRef Expression
1 sb6x.1 . . 3 xφ
21sbf 2026 . 2 ([y / x]φφ)
3 biidd 228 . . 3 (x = y → (φφ))
41, 3equsal 1960 . 2 (x(x = yφ) ↔ φ)
52, 4bitr4i 243 1 ([y / x]φx(x = yφ))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176  ∀wal 1540  Ⅎwnf 1544  [wsb 1648 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925 This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649 This theorem is referenced by: (None)
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