Theorem sbi1 2075

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		df-sb 2069	. 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 1819	. . . . 5 ⊢ (∀𝑥(𝑥 = 𝑧 → (𝜑 → 𝜓)) → (∀𝑥(𝑥 = 𝑧 → 𝜑) → ∀𝑥(𝑥 = 𝑧 → 𝜓)))
4	3	imim3i 64	. . . 4 ⊢ ((𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → (𝜑 → 𝜓))) → ((𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜑)) → (𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜓))))
5	4	al2imi 1819	. . 3 ⊢ (∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → (𝜑 → 𝜓))) → (∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜑)) → ∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜓))))
6		df-sb 2069	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜑)))
7		df-sb 2069	. . 3 ⊢ ([𝑦 / 𝑥]𝜓 ↔ ∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → 𝜓)))
8	5, 6, 7	3imtr4g 295	. 2 ⊢ (∀𝑧(𝑧 = 𝑦 → ∀𝑥(𝑥 = 𝑧 → (𝜑 → 𝜓))) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
9	1, 8	sylbi 216	1 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4  ∀wal 1537  [wsb 2068

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813

This theorem depends on definitions:  df-bi 206  df-sb 2069

This theorem is referenced by:  spsbim  2076  sbimi  2078  sb2imi  2079  sbim  2303  sbcim1  3767  2sb5ndVD  42419  2sb5ndALT  42441