Theorem bj-sbimedh 9911

Description: A strengthening of sbiedh 1670 (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 → (𝜓 → 𝜒)))

Assertion

Ref	Expression
bj-sbimedh	⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 → 𝜒))

Proof of Theorem bj-sbimedh

Step	Hyp	Ref	Expression
1		sb1 1649	. . 3 ⊢ ([𝑦 / 𝑥]𝜓 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))
2		bj-sbimedh.1	. . . 4 ⊢ (𝜑 → ∀𝑥𝜑)
3		bj-sbimedh.3	. . . . 5 ⊢ (𝜑 → (𝑥 = 𝑦 → (𝜓 → 𝜒)))
4	3	impd 242	. . . 4 ⊢ (𝜑 → ((𝑥 = 𝑦 ∧ 𝜓) → 𝜒))
5	2, 4	eximdh 1502	. . 3 ⊢ (𝜑 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜓) → ∃𝑥𝜒))
6	1, 5	syl5 28	. 2 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 → ∃𝑥𝜒))
7		bj-sbimedh.2	. . 3 ⊢ (𝜑 → (𝜒 → ∀𝑥𝜒))
8	2, 7	19.9hd 1552	. 2 ⊢ (𝜑 → (∃𝑥𝜒 → 𝜒))
9	6, 8	syld 40	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 → 𝜒))

Syntax hints:   → wi 4   ∧ wa 97  ∀wal 1241  ∃wex 1381  [wsb 1645

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 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-4 1400  ax-ial 1427

This theorem depends on definitions:  df-bi 110  df-sb 1646

This theorem is referenced by:  bj-sbimeh  9912