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Theorem sb6a 1861
Description: Equivalence for substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sb6a ([y / x]φx(x = y → [x / y]φ))
Distinct variable group:   x,y
Allowed substitution hints:   φ(x,y)

Proof of Theorem sb6a
StepHypRef Expression
1 sb6 1763 . 2 ([y / x]φx(x = yφ))
2 sbequ12 1651 . . . . 5 (y = x → (φ ↔ [x / y]φ))
32equcoms 1591 . . . 4 (x = y → (φ ↔ [x / y]φ))
43pm5.74i 169 . . 3 ((x = yφ) ↔ (x = y → [x / y]φ))
54albii 1356 . 2 (x(x = yφ) ↔ x(x = y → [x / y]φ))
61, 5bitri 173 1 ([y / x]φx(x = y → [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-8 1392  ax-11 1394  ax-4 1397  ax-17 1416  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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