Theorem sbal 2127

Description: Move universal quantifier in and out of substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sbal	⊢ ([z / y]∀xφ ↔ ∀x[z / y]φ)

Distinct variable groups:   x,y   x,z

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

Proof of Theorem sbal

Step	Hyp	Ref	Expression
1		a16gb 2050	. . . . 5 ⊢ (∀x x = z → (φ ↔ ∀xφ))
2	1	sbimi 1652	. . . 4 ⊢ ([z / y]∀x x = z → [z / y](φ ↔ ∀xφ))
3		sbequ5 2031	. . . 4 ⊢ ([z / y]∀x x = z ↔ ∀x x = z)
4		sbbi 2071	. . . 4 ⊢ ([z / y](φ ↔ ∀xφ) ↔ ([z / y]φ ↔ [z / y]∀xφ))
5	2, 3, 4	3imtr3i 256	. . 3 ⊢ (∀x x = z → ([z / y]φ ↔ [z / y]∀xφ))
6		a16gb 2050	. . 3 ⊢ (∀x x = z → ([z / y]φ ↔ ∀x[z / y]φ))
7	5, 6	bitr3d 246	. 2 ⊢ (∀x x = z → ([z / y]∀xφ ↔ ∀x[z / y]φ))
8		sbal1 2126	. 2 ⊢ (¬ ∀x x = z → ([z / y]∀xφ ↔ ∀x[z / y]φ))
9	7, 8	pm2.61i 156	1 ⊢ ([z / y]∀xφ ↔ ∀x[z / y]φ)

This proof depends on syntax axioms:   ↔ wb 176  ∀wal 1540  [wsb 1648

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  df-sb 1649

This theorem is used by:  sbex  2128  sbalv  2129  sbcal  3094  sbcalg  3095