Theorem sbn1 2105

Description: One direction of sbn 2279, using fewer axioms. Compare 19.2 1974. (Contributed by Steven Nguyen, 18-Aug-2023.)

Assertion

Ref	Expression
sbn1	⊢ ([𝑡 / 𝑥] ¬ 𝜑 → ¬ [𝑡 / 𝑥]𝜑)

Proof of Theorem sbn1

Step	Hyp	Ref	Expression
1		nsb 2104	. . 3 ⊢ (∀𝑥 ¬ ⊥ → ¬ [𝑡 / 𝑥]⊥)
2		fal 1551	. . 3 ⊢ ¬ ⊥
3	1, 2	mpg 1794	. 2 ⊢ ¬ [𝑡 / 𝑥]⊥
4		pm2.21 123	. . 3 ⊢ (¬ 𝜑 → (𝜑 → ⊥))
5	4	sb2imi 2073	. 2 ⊢ ([𝑡 / 𝑥] ¬ 𝜑 → ([𝑡 / 𝑥]𝜑 → [𝑡 / 𝑥]⊥))
6	3, 5	mtoi 199	1 ⊢ ([𝑡 / 𝑥] ¬ 𝜑 → ¬ [𝑡 / 𝑥]𝜑)

Syntax hints:  ¬ wn 3   → wi 4  ⊥wfal 1549  [wsb 2062

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965

This theorem depends on definitions:  df-bi 207  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063

This theorem is referenced by:  bj-ab0  36891