Metamath Proof Explorer


Theorem dral1ALT

Description: Alternate proof of dral1 , shorter but requiring ax-11 . (Contributed by NM, 24-Nov-1994) (Proof shortened by Wolf Lammen, 22-Apr-2018) (New usage is discouraged.) (Proof modification is discouraged.)

Ref Expression
Hypothesis dral1.1 ⊢ ∀ x x = y → φ ↔ ψ
Assertion dral1ALT ⊢ ∀ x x = y → ∀ x φ ↔ ∀ y ψ

Proof

Step Hyp Ref Expression
1 dral1.1 ⊢ ∀ x x = y → φ ↔ ψ
2 1 dral2 ⊢ ∀ x x = y → ∀ x φ ↔ ∀ x ψ
3 axc11 ⊢ ∀ x x = y → ∀ x ψ → ∀ y ψ
4 axc11r ⊢ ∀ x x = y → ∀ y ψ → ∀ x ψ
5 3 4 impbid ⊢ ∀ x x = y → ∀ x ψ ↔ ∀ y ψ
6 2 5 bitrd ⊢ ∀ x x = y → ∀ x φ ↔ ∀ y ψ