Theorem spvv 1963

Description: Version of spv 1913 with a disjoint variable condition. (Contributed by BJ, 31-May-2019.)

Hypothesis

Ref	Expression
spvv.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
spvv	⊢ (∀𝑥𝜑 → 𝜓)

Distinct variable groups:   𝑥,𝑦   𝜓,𝑥

Allowed substitution hints:   𝜑(𝑥, 𝑦)   𝜓(𝑦)

Proof of Theorem spvv

Step	Hyp	Ref	Expression
1		spvv.1	. 2 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	spv 1913	1 ⊢ (∀𝑥𝜑 → 𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 105  ∀wal 1400

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587

This proof depends on definitions:  df-bi 117  df-nf 1514

This theorem is used by:  chvarvv  1964