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Theorem sbal2 2020
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
StepHypRef Expression
1 hbnae 1721 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑦 ¬ ∀𝑥 𝑥 = 𝑦)
2 dveeq1 2019 . . . . . . 7 (¬ ∀𝑥 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
32alimi 1455 . . . . . 6 (∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
43hbnaes 1723 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
5 19.21ht 1581 . . . . 5 (∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧) → (∀𝑥(𝑦 = 𝑧𝜑) ↔ (𝑦 = 𝑧 → ∀𝑥𝜑)))
64, 5syl 14 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑦 = 𝑧𝜑) ↔ (𝑦 = 𝑧 → ∀𝑥𝜑)))
71, 6albidh 1480 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦𝑥(𝑦 = 𝑧𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑)))
8 alcom 1478 . . 3 (∀𝑦𝑥(𝑦 = 𝑧𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑))
97, 8bitr3di 195 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑)))
10 sb6 1886 . 2 ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
11 sb6 1886 . . 3 ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧𝜑))
1211albii 1470 . 2 (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑))
139, 10, 123bitr4g 223 1 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wb 105  wal 1351  [wsb 1762
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-fal 1359  df-nf 1461  df-sb 1763
This theorem is referenced by: (None)
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