Theorem sbimd 2281

Description: Deduction substituting both sides of an implication. (Contributed by Wolf Lammen, 24-Nov-2022.) Revise df-sb 2092. (Revised by Steven Nguyen, 9-Jul-2023.)

Hypotheses

Ref	Expression
sbimd.1	⊢ Ⅎ𝑥𝜑
sbimd.2	⊢ (𝜑 → (𝜓 → 𝜒))

Assertion

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

Proof of Theorem sbimd

Step	Hyp	Ref	Expression
1		sbimd.1	. . 3 ⊢ Ⅎ𝑥𝜑
2		sbimd.2	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3	1, 2	alrimi 2249	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒))
4		spsbim 2106	. 2 ⊢ (∀𝑥(𝜓 → 𝜒) → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜒))
5	3, 4	syl 17	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜒))

Syntax hints:   → wi 4  ∀wal 1559  Ⅎwnf 1804  [wsb 2091

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1816  ax-4 1830  ax-5 1931  ax-6 1988  ax-7 2029  ax-12 2213

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1801  df-nf 1805  df-sb 2092

This theorem is referenced by: (None)