Theorem spsbim 1836

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 55	. . . 4 ⊢ ((𝜑 → 𝜓) → ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜓)))
2	1	sps 1530	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜓)))
3		id 19	. . . . . 6 ⊢ ((𝜑 → 𝜓) → (𝜑 → 𝜓))
4	3	anim2d 335	. . . . 5 ⊢ ((𝜑 → 𝜓) → ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 ∧ 𝜓)))
5	4	alimi 1448	. . . 4 ⊢ (∀𝑥(𝜑 → 𝜓) → ∀𝑥((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 ∧ 𝜓)))
6		exim 1592	. . . 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 333	. 2 ⊢ (∀𝑥(𝜑 → 𝜓) → (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → ((𝑥 = 𝑦 → 𝜓) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))))
9		df-sb 1756	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
10		df-sb 1756	. 2 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ((𝑥 = 𝑦 → 𝜓) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓)))
11	8, 9, 10	3imtr4g 204	1 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ∧ wa 103  ∀wal 1346  ∃wex 1485  [wsb 1755

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1440  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-4 1503  ax-ial 1527

This theorem depends on definitions:  df-bi 116  df-sb 1756

This theorem is referenced by:  spsbbi  1837  hbsb4t  2006  moim  2083