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Theorem sbidm 1256
Description: An idempotent law for substitution.
Assertion
Ref Expression
sbidm ([y / x][y / x]φ ↔ [y / x]φ)

Proof of Theorem sbidm
StepHypRef Expression
1 sbequ12 1183 . . . 4 (x = y → ([y / x]φ ↔ [y / x][y / x]φ))
21bicomd 523 . . 3 (x = y → ([y / x][y / x]φ ↔ [y / x]φ))
32a4s 986 . 2 (x x = y → ([y / x][y / x]φ ↔ [y / x]φ))
4 hbnae 1149 . . 3 x x = yx ¬ x x = y)
5 hbsb2 1229 . . 3 x x = y → ([y / x]φx[y / x]φ))
6 pm4.2d 171 . . . 4 (x = y → ([y / x]φ ↔ [y / x]φ))
76a1i 8 . . 3 x x = y → (x = y → ([y / x]φ ↔ [y / x]φ)))
84, 5, 7sbied 1197 . 2 x x = y → ([y / x][y / x]φ ↔ [y / x]φ))
93, 8pm2.61i 126 1 ([y / x][y / x]φ ↔ [y / x]φ)
Colors of variables: wff set class
Syntax hints:  ¬ wn 2   → wi 3   ↔ wb 146  wal 956  [wsbc 1172
This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-10 968  ax-12 970  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-11o 1220
This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174
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