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Theorem bj-sbimedh 11101
Description: A strengthening of sbiedh 1714 (same proof). (Contributed by BJ, 16-Dec-2019.)
Hypotheses
Ref Expression
bj-sbimedh.1 (𝜑 → ∀𝑥𝜑)
bj-sbimedh.2 (𝜑 → (𝜒 → ∀𝑥𝜒))
bj-sbimedh.3 (𝜑 → (𝑥 = 𝑦 → (𝜓𝜒)))
Assertion
Ref Expression
bj-sbimedh (𝜑 → ([𝑦 / 𝑥]𝜓𝜒))

Proof of Theorem bj-sbimedh
StepHypRef Expression
1 sb1 1693 . . 3 ([𝑦 / 𝑥]𝜓 → ∃𝑥(𝑥 = 𝑦𝜓))
2 bj-sbimedh.1 . . . 4 (𝜑 → ∀𝑥𝜑)
3 bj-sbimedh.3 . . . . 5 (𝜑 → (𝑥 = 𝑦 → (𝜓𝜒)))
43impd 251 . . . 4 (𝜑 → ((𝑥 = 𝑦𝜓) → 𝜒))
52, 4eximdh 1545 . . 3 (𝜑 → (∃𝑥(𝑥 = 𝑦𝜓) → ∃𝑥𝜒))
61, 5syl5 32 . 2 (𝜑 → ([𝑦 / 𝑥]𝜓 → ∃𝑥𝜒))
7 bj-sbimedh.2 . . 3 (𝜑 → (𝜒 → ∀𝑥𝜒))
82, 719.9hd 1595 . 2 (𝜑 → (∃𝑥𝜒𝜒))
96, 8syld 44 1 (𝜑 → ([𝑦 / 𝑥]𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wal 1285  wex 1424  [wsb 1689
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1379  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-4 1443  ax-ial 1470
This theorem depends on definitions:  df-bi 115  df-sb 1690
This theorem is referenced by:  bj-sbimeh  11102
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