Theorem sbi1 2107

Description: Distribute substitution over implication. (Contributed by NM, 14-May-1993.) Remove dependencies on axioms. (Revised by Steven Nguyen, 24-Jul-2023.)

Assertion

Ref	Expression
sbi1	⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))

Proof of Theorem sbi1

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsb 2100	. 2 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ ∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → (𝜑 → 𝜓))))
2		ax-2 7	. . . . . 6 ⊢ ((𝑥 = 𝑧 → (𝜑 → 𝜓)) → ((𝑥 = 𝑧 → 𝜑) → (𝑥 = 𝑧 → 𝜓)))
3	2	al2imi 1847	. . . . 5 ⊢ (∀𝑥(𝑥 = 𝑧 → (𝜑 → 𝜓)) → (∀𝑥(𝑥 = 𝑧 → 𝜑) → ∀𝑥(𝑥 = 𝑧 → 𝜓)))
4	3	imim3i 65	. . . 4 ⊢ ((𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → (𝜑 → 𝜓))) → ((𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜑)) → (𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜓))))
5	4	al2imi 1847	. . 3 ⊢ (∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → (𝜑 → 𝜓))) → (∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜑)) → ∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜓))))
6		dfsb 2100	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜑)))
7		dfsb 2100	. . 3 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜓)))
8	5, 6, 7	3imtr4g 299	. 2 ⊢ (∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → (𝜑 → 𝜓))) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
9	1, 8	sylbi 220	1 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))

This proof depends on syntax axioms:   → wi 4  ∀wal 1567  [wsb 2098

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099

This theorem is used by:  spsbim  2108  sbimi  2110  sb2imi  2111  sbrimvw  2127  sbim  2336  sbcim1  3791  2sb5ndVD  45934  2sb5ndALT  45956