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Theorem bj-sbimedh 10298
Description: A strengthening of sbiedh 1686 (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 1665 . . 3 ([𝑦 / 𝑥]𝜓 → ∃𝑥(𝑥 = 𝑦𝜓))
2 bj-sbimedh.1 . . . 4 (𝜑 → ∀𝑥𝜑)
3 bj-sbimedh.3 . . . . 5 (𝜑 → (𝑥 = 𝑦 → (𝜓𝜒)))
43impd 246 . . . 4 (𝜑 → ((𝑥 = 𝑦𝜓) → 𝜒))
52, 4eximdh 1518 . . 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 1568 . 2 (𝜑 → (∃𝑥𝜒𝜒))
96, 8syld 44 1 (𝜑 → ([𝑦 / 𝑥]𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  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-ial 1443
This theorem depends on definitions:  df-bi 114  df-sb 1662
This theorem is referenced by:  bj-sbimeh  10299
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