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Theorem sbel2x 1852
 Description: Elimination of double substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbel2x (φxy((x = z y = w) [y / w][x / z]φ))
Distinct variable groups:   x,y,z   y,w   φ,x,y
Allowed substitution hints:   φ(z,w)

Proof of Theorem sbel2x
StepHypRef Expression
1 sbelx 1851 . . . . 5 ([x / z]φy(y = w [y / w][x / z]φ))
21anbi2i 433 . . . 4 ((x = z [x / z]φ) ↔ (x = z y(y = w [y / w][x / z]φ)))
32exbii 1474 . . 3 (x(x = z [x / z]φ) ↔ x(x = z y(y = w [y / w][x / z]φ)))
4 sbelx 1851 . . 3 (φx(x = z [x / z]φ))
5 exdistr 1765 . . 3 (xy(x = z (y = w [y / w][x / z]φ)) ↔ x(x = z y(y = w [y / w][x / z]φ)))
63, 4, 53bitr4i 201 . 2 (φxy(x = z (y = w [y / w][x / z]φ)))
7 anass 383 . . 3 (((x = z y = w) [y / w][x / z]φ) ↔ (x = z (y = w [y / w][x / z]φ)))
872exbii 1475 . 2 (xy((x = z y = w) [y / w][x / z]φ) ↔ xy(x = z (y = w [y / w][x / z]φ)))
96, 8bitr4i 176 1 (φxy((x = z y = w) [y / w][x / z]φ))
 Colors of variables: wff set class Syntax hints:   ∧ wa 97   ↔ wb 98  ∃wex 1358  [wsb 1623 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 1312  ax-gen 1314  ax-ie1 1359  ax-ie2 1360  ax-8 1372  ax-11 1374  ax-4 1377  ax-17 1396  ax-i9 1400  ax-ial 1405 This theorem depends on definitions:  df-bi 110  df-sb 1624 This theorem is referenced by: (None)
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