Theorem sbrim 2315

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 2152. (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		bi2.04 388	. . . . . . 7 ⊢ ((𝑥 = 𝑡 → (𝜑 → 𝜓)) ↔ (𝜑 → (𝑥 = 𝑡 → 𝜓)))
2	1	albii 1826	. . . . . 6 ⊢ (∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓)) ↔ ∀𝑥(𝜑 → (𝑥 = 𝑡 → 𝜓)))
3		sbrim.1	. . . . . . 7 ⊢ Ⅎ𝑥𝜑
4	3	19.21 2219	. . . . . 6 ⊢ (∀𝑥(𝜑 → (𝑥 = 𝑡 → 𝜓)) ↔ (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜓)))
5	2, 4	bitri 276	. . . . 5 ⊢ (∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓)) ↔ (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜓)))
6	5	imbi2i 337	. . . 4 ⊢ ((𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓))) ↔ (𝑡 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
7		bi2.04 388	. . . 4 ⊢ ((𝑡 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜓))) ↔ (𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
8	6, 7	bitri 276	. . 3 ⊢ ((𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓))) ↔ (𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
9	8	albii 1826	. 2 ⊢ (∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓))) ↔ ∀𝑡(𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
10		dfsb 2075	. 2 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ ∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → (𝜑 → 𝜓))))
11		dfsb 2075	. . . 4 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓)))
12	11	imbi2i 337	. . 3 ⊢ ((𝜑 → [𝑦 / 𝑥]𝜓) ↔ (𝜑 → ∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
13		19.21v 1946	. . 3 ⊢ (∀𝑡(𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))) ↔ (𝜑 → ∀𝑡(𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
14	12, 13	bitr4i 279	. 2 ⊢ ((𝜑 → [𝑦 / 𝑥]𝜓) ↔ ∀𝑡(𝜑 → (𝑡 = 𝑦 → ∀𝑥(𝑥 = 𝑡 → 𝜓))))
15	9, 10, 14	3bitr4i 304	1 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ↔ wb 207  ∀wal 1545  Ⅎwnf 1790  [wsb 2073

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-12 2189

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-nf 1791  df-sb 2074

This theorem is referenced by:  sbiedw  2325  sbied  2511  sbco2d  2520  2mosOLD  2654