Theorem sbi2v 1915

Description: Reverse direction of sbimv 1916. (Contributed by Jim Kingdon, 18-Jan-2018.)

Assertion

Ref	Expression
sbi2v	⊢ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) → [𝑦 / 𝑥](𝜑 → 𝜓))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)

Proof of Theorem sbi2v

Step	Hyp	Ref	Expression
1		19.38 1698	. . 3 ⊢ ((∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓)) → ∀𝑥((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜓)))
2		pm3.3 261	. . . . 5 ⊢ (((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜓)) → (𝑥 = 𝑦 → (𝜑 → (𝑥 = 𝑦 → 𝜓))))
3		pm2.04 82	. . . . 5 ⊢ ((𝜑 → (𝑥 = 𝑦 → 𝜓)) → (𝑥 = 𝑦 → (𝜑 → 𝜓)))
4	2, 3	syli 37	. . . 4 ⊢ (((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜓)) → (𝑥 = 𝑦 → (𝜑 → 𝜓)))
5	4	alimi 1477	. . 3 ⊢ (∀𝑥((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜓)) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)))
6	1, 5	syl 14	. 2 ⊢ ((∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓)) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)))
7		sb5 1910	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
8		sb6 1909	. . 3 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
9	7, 8	imbi12i 239	. 2 ⊢ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ↔ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓)))
10		sb6 1909	. 2 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)))
11	6, 9, 10	3imtr4i 201	1 ⊢ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) → [𝑦 / 𝑥](𝜑 → 𝜓))

Syntax hints:   → wi 4   ∧ wa 104  ∀wal 1370  ∃wex 1514  [wsb 1784

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1469  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-11 1528  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557

This theorem depends on definitions:  df-bi 117  df-sb 1785

This theorem is referenced by:  sbimv  1916