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Theorem sbal2 1975
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 alcom 1439 . . 3 (∀𝑦𝑥(𝑦 = 𝑧𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑))
2 hbnae 1684 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑦 ¬ ∀𝑥 𝑥 = 𝑦)
3 dveeq1 1972 . . . . . . 7 (¬ ∀𝑥 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
43alimi 1416 . . . . . 6 (∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
54hbnaes 1686 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
6 19.21ht 1545 . . . . 5 (∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧) → (∀𝑥(𝑦 = 𝑧𝜑) ↔ (𝑦 = 𝑧 → ∀𝑥𝜑)))
75, 6syl 14 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑦 = 𝑧𝜑) ↔ (𝑦 = 𝑧 → ∀𝑥𝜑)))
82, 7albidh 1441 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦𝑥(𝑦 = 𝑧𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑)))
91, 8syl5rbbr 194 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑)))
10 sb6 1842 . 2 ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
11 sb6 1842 . . 3 ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧𝜑))
1211albii 1431 . 2 (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑))
139, 10, 123bitr4g 222 1 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wb 104  wal 1314  [wsb 1720
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 588  ax-in2 589  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500
This theorem depends on definitions:  df-bi 116  df-tru 1319  df-fal 1322  df-nf 1422  df-sb 1721
This theorem is referenced by: (None)
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