Theorem sbal2 1005

Description: Move quantifier in and out of substitution.

Assertion

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

Distinct variable group(s):   y,z   x,z

Proof of Theorem sbal2

Step	Hyp	Ref	Expression
1		eq6 826	. . . 4 ⊢ (¬ ∀x x = y → ∀y ¬ ∀x x = y)
2		ddeeq1 1001	. . . . . . 7 ⊢ (¬ ∀x x = y → (y = z → ∀x y = z))
3	2	19.20i 691	. . . . . 6 ⊢ (∀x ¬ ∀x x = y → ∀x(y = z → ∀x y = z))
4	3	eq6s 827	. . . . 5 ⊢ (¬ ∀x x = y → ∀x(y = z → ∀x y = z))
5		19.21g 792	. . . . 5 ⊢ (∀x(y = z → ∀x y = z) → (∀x(y = z → φ) ↔ (y = z → ∀xφ)))
6	4, 5	syl 12	. . . 4 ⊢ (¬ ∀x x = y → (∀x(y = z → φ) ↔ (y = z → ∀xφ)))
7	1, 6	biald 782	. . 3 ⊢ (¬ ∀x x = y → (∀y∀x(y = z → φ) ↔ ∀y(y = z → ∀xφ)))
8		alcom 715	. . 3 ⊢ (∀y∀x(y = z → φ) ↔ ∀x∀y(y = z → φ))
9	7, 8	syl5rbbr 413	. 2 ⊢ (¬ ∀x x = y → (∀y(y = z → ∀xφ) ↔ ∀x∀y(y = z → φ)))
10		sb6 989	. 2 ⊢ ([z / y]∀xφ ↔ ∀y(y = z → ∀xφ))
11		sb6 989	. . 3 ⊢ ([z / y]φ ↔ ∀y(y = z → φ))
12	11	bial 695	. 2 ⊢ (∀x[z / y]φ ↔ ∀x∀y(y = z → φ))
13	9, 10, 12	3bitr4g 428	1 ⊢ (¬ ∀x x = y → ([z / y]∀xφ ↔ ∀x[z / y]φ))

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127  ∀wal 672   = weq 797  [wsb 852

This theorem is referenced by:  axrepndlem2 3739

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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