Theorem bj-sblem 37507

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

Assertion

Ref	Expression
bj-sblem	⊢ (∀𝑥(𝜑 → (𝜓 ↔ 𝜒)) → (∀𝑥(𝜑 → 𝜓) ↔ (∃𝑥𝜑 → 𝜒)))

Distinct variable group:   𝜒,𝑥

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

Proof of Theorem bj-sblem

Step	Hyp	Ref	Expression
1		pm5.74 273	. . . 4 ⊢ ((𝜑 → (𝜓 ↔ 𝜒)) ↔ ((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)))
2	1	albii 1848	. . 3 ⊢ (∀𝑥(𝜑 → (𝜓 ↔ 𝜒)) ↔ ∀𝑥((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)))
3		albi 1847	. . 3 ⊢ (∀𝑥((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)) → (∀𝑥(𝜑 → 𝜓) ↔ ∀𝑥(𝜑 → 𝜒)))
4	2, 3	sylbi 220	. 2 ⊢ (∀𝑥(𝜑 → (𝜓 ↔ 𝜒)) → (∀𝑥(𝜑 → 𝜓) ↔ ∀𝑥(𝜑 → 𝜒)))
5		19.23v 1971	. 2 ⊢ (∀𝑥(𝜑 → 𝜒) ↔ (∃𝑥𝜑 → 𝜒))
6	4, 5	bitrdi 290	1 ⊢ (∀𝑥(𝜑 → (𝜓 ↔ 𝜒)) → (∀𝑥(𝜑 → 𝜓) ↔ (∃𝑥𝜑 → 𝜒)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567  ∃wex 1808

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

This proof depends on definitions:  df-bi 210  df-ex 1809

This theorem is used by:  bj-sbievw  37510