Theorem sbiedh 1670

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbieh 1673). New proofs should use sbied 1671 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 → (𝜓 ↔ 𝜒)))

Assertion

Ref	Expression
sbiedh	⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ 𝜒))

Proof of Theorem sbiedh

Step	Hyp	Ref	Expression
1		sb1 1649	. . . 4 ⊢ ([𝑦 / 𝑥]𝜓 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))
2		sbiedh.1	. . . . 5 ⊢ (𝜑 → ∀𝑥𝜑)
3		sbiedh.3	. . . . . . 7 ⊢ (𝜑 → (𝑥 = 𝑦 → (𝜓 ↔ 𝜒)))
4		bi1 111	. . . . . . 7 ⊢ ((𝜓 ↔ 𝜒) → (𝜓 → 𝜒))
5	3, 4	syl6 29	. . . . . 6 ⊢ (𝜑 → (𝑥 = 𝑦 → (𝜓 → 𝜒)))
6	5	impd 242	. . . . 5 ⊢ (𝜑 → ((𝑥 = 𝑦 ∧ 𝜓) → 𝜒))
7	2, 6	eximdh 1502	. . . 4 ⊢ (𝜑 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜓) → ∃𝑥𝜒))
8	1, 7	syl5 28	. . 3 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 → ∃𝑥𝜒))
9		sbiedh.2	. . . 4 ⊢ (𝜑 → (𝜒 → ∀𝑥𝜒))
10	2, 9	19.9hd 1552	. . 3 ⊢ (𝜑 → (∃𝑥𝜒 → 𝜒))
11	8, 10	syld 40	. 2 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 → 𝜒))
12		bi2 121	. . . . . . 7 ⊢ ((𝜓 ↔ 𝜒) → (𝜒 → 𝜓))
13	3, 12	syl6 29	. . . . . 6 ⊢ (𝜑 → (𝑥 = 𝑦 → (𝜒 → 𝜓)))
14	13	com23 72	. . . . 5 ⊢ (𝜑 → (𝜒 → (𝑥 = 𝑦 → 𝜓)))
15	2, 14	alimdh 1356	. . . 4 ⊢ (𝜑 → (∀𝑥𝜒 → ∀𝑥(𝑥 = 𝑦 → 𝜓)))
16		sb2 1650	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) → [𝑦 / 𝑥]𝜓)
17	15, 16	syl6 29	. . 3 ⊢ (𝜑 → (∀𝑥𝜒 → [𝑦 / 𝑥]𝜓))
18	9, 17	syld 40	. 2 ⊢ (𝜑 → (𝜒 → [𝑦 / 𝑥]𝜓))
19	11, 18	impbid 120	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  ∀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-i9 1423  ax-ial 1427

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

This theorem is referenced by:  sbied  1671  sbieh  1673  sbcomxyyz  1846