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Theorem sbiedh 1667
Description: Conversion of implicit substitution to explicit substitution (deduction version of sbieh 1670). New proofs should use sbied 1668 instead. (Contributed by NM, 30-Jun-1994.) (Proof shortened by Andrew Salmon, 25-May-2011.) (New usage is discouraged.)
Hypotheses
Ref Expression
sbiedh.1 (φxφ)
sbiedh.2 (φ → (χxχ))
sbiedh.3 (φ → (x = y → (ψχ)))
Assertion
Ref Expression
sbiedh (φ → ([y / x]ψχ))

Proof of Theorem sbiedh
StepHypRef Expression
1 sb1 1646 . . . 4 ([y / x]ψx(x = y ψ))
2 sbiedh.1 . . . . 5 (φxφ)
3 sbiedh.3 . . . . . . 7 (φ → (x = y → (ψχ)))
4 bi1 111 . . . . . . 7 ((ψχ) → (ψχ))
53, 4syl6 29 . . . . . 6 (φ → (x = y → (ψχ)))
65impd 242 . . . . 5 (φ → ((x = y ψ) → χ))
72, 6eximdh 1499 . . . 4 (φ → (x(x = y ψ) → xχ))
81, 7syl5 28 . . 3 (φ → ([y / x]ψxχ))
9 sbiedh.2 . . . 4 (φ → (χxχ))
102, 919.9hd 1549 . . 3 (φ → (xχχ))
118, 10syld 40 . 2 (φ → ([y / x]ψχ))
12 bi2 121 . . . . . . 7 ((ψχ) → (χψ))
133, 12syl6 29 . . . . . 6 (φ → (x = y → (χψ)))
1413com23 72 . . . . 5 (φ → (χ → (x = yψ)))
152, 14alimdh 1353 . . . 4 (φ → (xχx(x = yψ)))
16 sb2 1647 . . . 4 (x(x = yψ) → [y / x]ψ)
1715, 16syl6 29 . . 3 (φ → (xχ → [y / x]ψ))
189, 17syld 40 . 2 (φ → (χ → [y / x]ψ))
1911, 18impbid 120 1 (φ → ([y / x]ψχ))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wb 98  wal 1240  wex 1378  [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-4 1397  ax-i9 1420  ax-ial 1424
This theorem depends on definitions:  df-bi 110  df-sb 1643
This theorem is referenced by:  sbied  1668  sbieh  1670  sbcomxyyz  1843
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