Theorem bj-sbievw2 36847

Description: Lemma for substitution. (Contributed by BJ, 23-Jul-2023.)

Assertion

Ref	Expression
bj-sbievw2	⊢ ([𝑦 / 𝑥](𝜓 → 𝜑) → (𝜓 → [𝑦 / 𝑥]𝜑))

Distinct variable groups:   𝜓,𝑥   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑦)

Proof of Theorem bj-sbievw2

Step	Hyp	Ref	Expression
1		sb6 2085	. 2 ⊢ ([𝑦 / 𝑥](𝜓 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → (𝜓 → 𝜑)))
2		bj-sblem2 36844	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → (𝜓 → 𝜑)) → ((∃𝑥 𝑥 = 𝑦 → 𝜓) → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
3		jarr 106	. . . 4 ⊢ (((∃𝑥 𝑥 = 𝑦 → 𝜓) → ∀𝑥(𝑥 = 𝑦 → 𝜑)) → (𝜓 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
4		sb6 2085	. . . 4 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
5	3, 4	imbitrrdi 252	. . 3 ⊢ (((∃𝑥 𝑥 = 𝑦 → 𝜓) → ∀𝑥(𝑥 = 𝑦 → 𝜑)) → (𝜓 → [𝑦 / 𝑥]𝜑))
6	2, 5	syl 17	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → (𝜓 → 𝜑)) → (𝜓 → [𝑦 / 𝑥]𝜑))
7	1, 6	sylbi 217	1 ⊢ ([𝑦 / 𝑥](𝜓 → 𝜑) → (𝜓 → [𝑦 / 𝑥]𝜑))

Syntax hints:   → wi 4  ∀wal 1538  ∃wex 1779  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-sb 2065

This theorem is referenced by: (None)