Theorem bj-sbievw1 37508

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

Assertion

Ref	Expression
bj-sbievw1	⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → 𝜓))

Distinct variable groups:   𝜓,𝑥   𝑥,𝑦

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

Proof of Theorem bj-sbievw1

Step	Hyp	Ref	Expression
1		sb6 2118	. 2 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)))
2		bj-sblem1 37505	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)) → (∀𝑥(𝑥 = 𝑦 → 𝜑) → (∃𝑥 𝑥 = 𝑦 → 𝜓)))
3		sb6 2118	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
4		ax6ev 1998	. . . 4 ⊢ ∃𝑥 𝑥 = 𝑦
5	4	a1bi 365	. . 3 ⊢ (𝜓 ↔ (∃𝑥 𝑥 = 𝑦 → 𝜓))
6	2, 3, 5	3imtr4g 299	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)) → ([𝑦 / 𝑥]𝜑 → 𝜓))
7	1, 6	sylbi 220	1 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → 𝜓))

This proof depends on syntax axioms:   → wi 4  ∀wal 1567  ∃wex 1808  [wsb 2095

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1809  df-sb 2096

This theorem is used by: (None)