Theorem spsbim 1724

Description: Specialization of implication. (Contributed by NM, 5-Aug-1993.) (Proof rewritten by Jim Kingdon, 21-Jan-2018.)

Assertion

Ref	Expression
spsbim	⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))

Proof of Theorem spsbim

Step	Hyp	Ref	Expression
1		imim2 49	. . . 4 ⊢ ((𝜑 → 𝜓) → ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜓)))
2	1	sps 1430	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜓)))
3		id 19	. . . . . 6 ⊢ ((𝜑 → 𝜓) → (𝜑 → 𝜓))
4	3	anim2d 320	. . . . 5 ⊢ ((𝜑 → 𝜓) → ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 ∧ 𝜓)))
5	4	alimi 1344	. . . 4 ⊢ (∀𝑥(𝜑 → 𝜓) → ∀𝑥((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 ∧ 𝜓)))
6		exim 1490	. . . 4 ⊢ (∀𝑥((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 ∧ 𝜓)) → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
7	5, 6	syl 14	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
8	2, 7	anim12d 318	. 2 ⊢ (∀𝑥(𝜑 → 𝜓) → (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → ((𝑥 = 𝑦 → 𝜓) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))))
9		df-sb 1646	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
10		df-sb 1646	. 2 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ((𝑥 = 𝑦 → 𝜓) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
11	8, 9, 10	3imtr4g 194	1 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))

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:  spsbbi  1725  hbsb4t  1889  moim  1964