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Theorem sbi1 2063
 Description: Removal of implication from substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbi1 ([y / x](φψ) → ([y / x]φ → [y / x]ψ))

Proof of Theorem sbi1
StepHypRef Expression
1 sbequ2 1650 . . . . 5 (x = y → ([y / x]φφ))
2 sbequ2 1650 . . . . 5 (x = y → ([y / x](φψ) → (φψ)))
31, 2syl5d 62 . . . 4 (x = y → ([y / x](φψ) → ([y / x]φψ)))
4 sbequ1 1918 . . . 4 (x = y → (ψ → [y / x]ψ))
53, 4syl6d 64 . . 3 (x = y → ([y / x](φψ) → ([y / x]φ → [y / x]ψ)))
65sps 1754 . 2 (x x = y → ([y / x](φψ) → ([y / x]φ → [y / x]ψ)))
7 sb4 2053 . . 3 x x = y → ([y / x]φx(x = yφ)))
8 sb4 2053 . . . 4 x x = y → ([y / x](φψ) → x(x = y → (φψ))))
9 ax-2 6 . . . . . 6 ((x = y → (φψ)) → ((x = yφ) → (x = yψ)))
109al2imi 1561 . . . . 5 (x(x = y → (φψ)) → (x(x = yφ) → x(x = yψ)))
11 sb2 2023 . . . . 5 (x(x = yψ) → [y / x]ψ)
1210, 11syl6 29 . . . 4 (x(x = y → (φψ)) → (x(x = yφ) → [y / x]ψ))
138, 12syl6 29 . . 3 x x = y → ([y / x](φψ) → (x(x = yφ) → [y / x]ψ)))
147, 13syl5d 62 . 2 x x = y → ([y / x](φψ) → ([y / x]φ → [y / x]ψ)))
156, 14pm2.61i 156 1 ([y / x](φψ) → ([y / x]φ → [y / x]ψ))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4  ∀wal 1540  [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:  sbim  2065  spsbim  2076
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