Theorem spv 1998

Description: Specialization, using implicit substitution. (Contributed by NM, 30-Aug-1993.)

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 198	. 2 ⊢ (x = y → (φ → ψ))
3	2	spimv 1990	1 ⊢ (∀xφ → ψ)

This proof depends on syntax axioms:   → wi 4   ↔ wb 176  ∀wal 1540

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This proof depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545

This theorem is used by:  chvarv  2013  ax10-16  2190  ru  3046  sfintfin  4533  spfinsfincl  4540