Theorem bj-equsal2 32000

Description: One direction of equsal 2279. (Contributed by BJ, 30-Sep-2018.)

Hypotheses

Ref	Expression
bj-equsal2.1	⊢ Ⅎ𝑥𝜑
bj-equsal2.2	⊢ (𝑥 = 𝑦 → (𝜑 → 𝜓))

Assertion

Ref	Expression
bj-equsal2	⊢ (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜓))

Proof of Theorem bj-equsal2

Step	Hyp	Ref	Expression
1		bj-equsal2.1	. . 3 ⊢ Ⅎ𝑥𝜑
2	1	bj-equsal1ti 31998	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜑)
3		bj-equsal2.2	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 → 𝜓))
4	3	a2i 14	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜓))
5	4	alimi 1730	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓))
6	2, 5	sylbir 224	1 ⊢ (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜓))

Syntax hints:   → wi 4  ∀wal 1473  Ⅎwnf 1699

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-12 2034  ax-13 2234

This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696  df-nf 1701

This theorem is referenced by:  bj-equsal  32001