Theorem spv 2427

Description: Specialization, using implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2406. Use the weaker spvv 2021 if possible. (Contributed by NM, 30-Aug-1993.) (New usage is discouraged.)

Hypothesis

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

Assertion

Ref	Expression
spv	⊢ (∀𝑥𝜑 → 𝜓)

Distinct variable group:   𝜓,𝑥

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

Proof of Theorem spv

Step	Hyp	Ref	Expression
1		spv.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	biimpd 232	. 2 ⊢ (𝑥 = 𝑦 → (𝜑 → 𝜓))
3	2	spimv 2424	1 ⊢ (∀𝑥𝜑 → 𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-12 2216  ax-13 2406

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817

This theorem is used by:  axc11n-16  39770