Theorem bj-sblem 34168

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		pm5.74 272	. . . 4 ⊢ ((𝜑 → (𝜓 ↔ 𝜒)) ↔ ((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)))
2	1	albii 1820	. . 3 ⊢ (∀𝑥(𝜑 → (𝜓 ↔ 𝜒)) ↔ ∀𝑥((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)))
3		albi 1819	. . 3 ⊢ (∀𝑥((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)) → (∀𝑥(𝜑 → 𝜓) ↔ ∀𝑥(𝜑 → 𝜒)))
4	2, 3	sylbi 219	. 2 ⊢ (∀𝑥(𝜑 → (𝜓 ↔ 𝜒)) → (∀𝑥(𝜑 → 𝜓) ↔ ∀𝑥(𝜑 → 𝜒)))
5		19.23v 1943	. 2 ⊢ (∀𝑥(𝜑 → 𝜒) ↔ (∃𝑥𝜑 → 𝜒))
6	4, 5	syl6bb 289	1 ⊢ (∀𝑥(𝜑 → (𝜓 ↔ 𝜒)) → (∀𝑥(𝜑 → 𝜓) ↔ (∃𝑥𝜑 → 𝜒)))

Syntax hints:   → wi 4   ↔ wb 208  ∀wal 1535  ∃wex 1780

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911

This theorem depends on definitions:  df-bi 209  df-ex 1781

This theorem is referenced by:  bj-sbievw  34171