Theorem bj-hbaeb

Description: Biconditional version of hbae . (Contributed by BJ, 6-Oct-2018) (Proof modification is discouraged.)

	Ref	Expression
Assertion	bj-hbaeb	⊢ ( ∀ 𝑥 𝑥 = 𝑦 ↔ ∀ 𝑧 ∀ 𝑥 𝑥 = 𝑦 )

Proof

Step	Hyp	Ref	Expression
1		bj-hbaeb2	⊢ ( ∀ 𝑥 𝑥 = 𝑦 ↔ ∀ 𝑥 ∀ 𝑧 𝑥 = 𝑦 )
2		alcom	⊢ ( ∀ 𝑥 ∀ 𝑧 𝑥 = 𝑦 ↔ ∀ 𝑧 ∀ 𝑥 𝑥 = 𝑦 )
3	1 2	bitri	⊢ ( ∀ 𝑥 𝑥 = 𝑦 ↔ ∀ 𝑧 ∀ 𝑥 𝑥 = 𝑦 )