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Theorem sbiedh 1686
Description: Conversion of implicit substitution to explicit substitution (deduction version of sbieh 1689). New proofs should use sbied 1687 instead. (Contributed by NM, 30-Jun-1994.) (Proof shortened by Andrew Salmon, 25-May-2011.) (New usage is discouraged.)
Hypotheses
Ref Expression
sbiedh.1 (𝜑 → ∀𝑥𝜑)
sbiedh.2 (𝜑 → (𝜒 → ∀𝑥𝜒))
sbiedh.3 (𝜑 → (𝑥 = 𝑦 → (𝜓𝜒)))
Assertion
Ref Expression
sbiedh (𝜑 → ([𝑦 / 𝑥]𝜓𝜒))

Proof of Theorem sbiedh
StepHypRef Expression
1 sb1 1665 . . . 4 ([𝑦 / 𝑥]𝜓 → ∃𝑥(𝑥 = 𝑦𝜓))
2 sbiedh.1 . . . . 5 (𝜑 → ∀𝑥𝜑)
3 sbiedh.3 . . . . . . 7 (𝜑 → (𝑥 = 𝑦 → (𝜓𝜒)))
4 bi1 115 . . . . . . 7 ((𝜓𝜒) → (𝜓𝜒))
53, 4syl6 33 . . . . . 6 (𝜑 → (𝑥 = 𝑦 → (𝜓𝜒)))
65impd 246 . . . . 5 (𝜑 → ((𝑥 = 𝑦𝜓) → 𝜒))
72, 6eximdh 1518 . . . 4 (𝜑 → (∃𝑥(𝑥 = 𝑦𝜓) → ∃𝑥𝜒))
81, 7syl5 32 . . 3 (𝜑 → ([𝑦 / 𝑥]𝜓 → ∃𝑥𝜒))
9 sbiedh.2 . . . 4 (𝜑 → (𝜒 → ∀𝑥𝜒))
102, 919.9hd 1568 . . 3 (𝜑 → (∃𝑥𝜒𝜒))
118, 10syld 44 . 2 (𝜑 → ([𝑦 / 𝑥]𝜓𝜒))
12 bi2 125 . . . . . . 7 ((𝜓𝜒) → (𝜒𝜓))
133, 12syl6 33 . . . . . 6 (𝜑 → (𝑥 = 𝑦 → (𝜒𝜓)))
1413com23 76 . . . . 5 (𝜑 → (𝜒 → (𝑥 = 𝑦𝜓)))
152, 14alimdh 1372 . . . 4 (𝜑 → (∀𝑥𝜒 → ∀𝑥(𝑥 = 𝑦𝜓)))
16 sb2 1666 . . . 4 (∀𝑥(𝑥 = 𝑦𝜓) → [𝑦 / 𝑥]𝜓)
1715, 16syl6 33 . . 3 (𝜑 → (∀𝑥𝜒 → [𝑦 / 𝑥]𝜓))
189, 17syld 44 . 2 (𝜑 → (𝜒 → [𝑦 / 𝑥]𝜓))
1911, 18impbid 124 1 (𝜑 → ([𝑦 / 𝑥]𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102  wal 1257  wex 1397  [wsb 1661
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-4 1416  ax-i9 1439  ax-ial 1443
This theorem depends on definitions:  df-bi 114  df-sb 1662
This theorem is referenced by:  sbied  1687  sbieh  1689  sbcomxyyz  1862
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