Theorem sbal 2169

Description: Move universal quantifier in and out of substitution. (Contributed by NM, 16-May-1993.) (Proof shortened by Wolf Lammen, 29-Sep-2018.) Reduce dependencies on axioms. (Revised by Steven Nguyen, 13-Aug-2023.)

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

Dummy variable 𝑤 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		alcom 2159	. 2 ⊢ (∀𝑤∀𝑥(𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)) ↔ ∀𝑥∀𝑤(𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)))
2		19.21v 1939	. . . . . . 7 ⊢ (∀𝑥(𝑦 = 𝑤 → 𝜑) ↔ (𝑦 = 𝑤 → ∀𝑥𝜑))
3	2	albii 1819	. . . . . 6 ⊢ (∀𝑦∀𝑥(𝑦 = 𝑤 → 𝜑) ↔ ∀𝑦(𝑦 = 𝑤 → ∀𝑥𝜑))
4		alcom 2159	. . . . . 6 ⊢ (∀𝑦∀𝑥(𝑦 = 𝑤 → 𝜑) ↔ ∀𝑥∀𝑦(𝑦 = 𝑤 → 𝜑))
5	3, 4	bitr3i 277	. . . . 5 ⊢ (∀𝑦(𝑦 = 𝑤 → ∀𝑥𝜑) ↔ ∀𝑥∀𝑦(𝑦 = 𝑤 → 𝜑))
6	5	imbi2i 336	. . . 4 ⊢ ((𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → ∀𝑥𝜑)) ↔ (𝑤 = 𝑧 → ∀𝑥∀𝑦(𝑦 = 𝑤 → 𝜑)))
7	6	albii 1819	. . 3 ⊢ (∀𝑤(𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → ∀𝑥𝜑)) ↔ ∀𝑤(𝑤 = 𝑧 → ∀𝑥∀𝑦(𝑦 = 𝑤 → 𝜑)))
8		df-sb 2065	. . 3 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑤(𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → ∀𝑥𝜑)))
9		19.21v 1939	. . . 4 ⊢ (∀𝑥(𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)) ↔ (𝑤 = 𝑧 → ∀𝑥∀𝑦(𝑦 = 𝑤 → 𝜑)))
10	9	albii 1819	. . 3 ⊢ (∀𝑤∀𝑥(𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)) ↔ ∀𝑤(𝑤 = 𝑧 → ∀𝑥∀𝑦(𝑦 = 𝑤 → 𝜑)))
11	7, 8, 10	3bitr4i 303	. 2 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑤∀𝑥(𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)))
12		df-sb 2065	. . 3 ⊢ ([𝑧 / 𝑦]𝜑 ↔ ∀𝑤(𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)))
13	12	albii 1819	. 2 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥∀𝑤(𝑤 = 𝑧 → ∀𝑦(𝑦 = 𝑤 → 𝜑)))
14	1, 11, 13	3bitr4i 303	1 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1538  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-11 2157

This theorem depends on definitions:  df-bi 207  df-ex 1780  df-sb 2065

This theorem is referenced by:  sbalv  2170  hbsbw  2171  sbnf  2312  sbnfOLD  2313  nfaba1  2913  sbralie  3358  sbcal  3849  ax11-pm2  36837  ichal  47453