Theorem sbrim 2305

Description: Substitution in an implication with a variable not free in the antecedent affects only the consequent. (Contributed by NM, 2-Jun-1993.) (Revised by Mario Carneiro, 4-Oct-2016.) Avoid ax-10 2142. (Revised by GG, 20-Nov-2024.)

Hypothesis

Ref	Expression
sbrim.1	⊢ Ⅎ𝑥𝜑

Assertion

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

Proof of Theorem sbrim

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		bi2.04 387	. . . . . . 7 ⊢ ((𝑥 = 𝑡 → (𝜑 → 𝜓)) ↔ (𝜑 → (𝑥 = 𝑡 → 𝜓)))
2	1	albii 1819	. . . . . 6 ⊢ (∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓)) ↔ ∀𝑥(𝜑 → (𝑥 = 𝑡 → 𝜓)))
3		sbrim.1	. . . . . . 7 ⊢ Ⅎ𝑥𝜑
4	3	19.21 2208	. . . . . 6 ⊢ (∀𝑥(𝜑 → (𝑥 = 𝑡 → 𝜓)) ↔ (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜓)))
5	2, 4	bitri 275	. . . . 5 ⊢ (∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓)) ↔ (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜓)))
6	5	imbi2i 336	. . . 4 ⊢ ((𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓))) ↔ (𝑡 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
7		bi2.04 387	. . . 4 ⊢ ((𝑡 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜓))) ↔ (𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
8	6, 7	bitri 275	. . 3 ⊢ ((𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓))) ↔ (𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
9	8	albii 1819	. 2 ⊢ (∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓))) ↔ ∀𝑡(𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
10		df-sb 2066	. 2 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ ∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓))))
11		df-sb 2066	. . . 4 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓)))
12	11	imbi2i 336	. . 3 ⊢ ((𝜑 → [𝑦 / 𝑥]𝜓) ↔ (𝜑 → ∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
13		19.21v 1939	. . 3 ⊢ (∀𝑡(𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))) ↔ (𝜑 → ∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
14	12, 13	bitr4i 278	. 2 ⊢ ((𝜑 → [𝑦 / 𝑥]𝜓) ↔ ∀𝑡(𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
15	9, 10, 14	3bitr4i 303	1 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1538  Ⅎwnf 1783  [wsb 2065

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-12 2178

This theorem depends on definitions:  df-bi 207  df-ex 1780  df-nf 1784  df-sb 2066

This theorem is referenced by:  sbiedw  2317  sbied  2508  sbco2d  2517  2mosOLD  2650