Theorem bj-equsal1 36006

Description: One direction of equsal 2415. (Contributed by BJ, 30-Sep-2018.)

Hypotheses

Ref	Expression
bj-equsal1.1	⊢ Ⅎ𝑥𝜓
bj-equsal1.2	⊢ (𝑥 = 𝑦 → (𝜑 → 𝜓))

Assertion

Ref	Expression
bj-equsal1	⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → 𝜓)

Proof of Theorem bj-equsal1

Step	Hyp	Ref	Expression
1		bj-equsal1.2	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 → 𝜓))
2	1	a2i 14	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜓))
3	2	alimi 1812	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓))
4		bj-equsal1.1	. . 3 ⊢ Ⅎ𝑥𝜓
5	4	bj-equsal1ti 36005	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ 𝜓)
6	3, 5	sylib 217	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → 𝜓)

Syntax hints:   → wi 4  ∀wal 1538  Ⅎwnf 1784

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 1912  ax-6 1970  ax-7 2010  ax-12 2170  ax-13 2370

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1781  df-nf 1785

This theorem is referenced by:  bj-equsal  36008