Theorem sbal2 2134

Description: Move quantifier in and out of substitution. (Contributed by NM, 2-Jan-2002.)

Assertion

Ref	Expression
sbal2	⊢ (¬ ∀x x = y → ([z / y]∀xφ ↔ ∀x[z / y]φ))

Distinct variable groups:   y,z   x,z

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

Proof of Theorem sbal2

Step	Hyp	Ref	Expression
1		alcom 1737	. . 3 ⊢ (∀y∀x(y = z → φ) ↔ ∀x∀y(y = z → φ))
2		nfnae 1956	. . . 4 ⊢ Ⅎy ¬ ∀x x = y
3		nfnae 1956	. . . . . 6 ⊢ Ⅎx ¬ ∀x x = y
4		dveeq1 2018	. . . . . 6 ⊢ (¬ ∀x x = y → (y = z → ∀x y = z))
5	3, 4	nfd 1766	. . . . 5 ⊢ (¬ ∀x x = y → Ⅎx y = z)
6		19.21t 1795	. . . . 5 ⊢ (Ⅎx y = z → (∀x(y = z → φ) ↔ (y = z → ∀xφ)))
7	5, 6	syl 15	. . . 4 ⊢ (¬ ∀x x = y → (∀x(y = z → φ) ↔ (y = z → ∀xφ)))
8	2, 7	albid 1772	. . 3 ⊢ (¬ ∀x x = y → (∀y∀x(y = z → φ) ↔ ∀y(y = z → ∀xφ)))
9	1, 8	syl5rbbr 251	. 2 ⊢ (¬ ∀x x = y → (∀y(y = z → ∀xφ) ↔ ∀x∀y(y = z → φ)))
10		sb6 2099	. 2 ⊢ ([z / y]∀xφ ↔ ∀y(y = z → ∀xφ))
11		sb6 2099	. . 3 ⊢ ([z / y]φ ↔ ∀y(y = z → φ))
12	11	albii 1566	. 2 ⊢ (∀x[z / y]φ ↔ ∀x∀y(y = z → φ))
13	9, 10, 12	3bitr4g 279	1 ⊢ (¬ ∀x x = y → ([z / y]∀xφ ↔ ∀x[z / y]φ))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176  ∀wal 1540  Ⅎwnf 1544  [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: (None)