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Theorem sbal2 1940
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 1408 . . 3 (∀𝑦𝑥(𝑦 = 𝑧𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑))
2 hbnae 1650 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑦 ¬ ∀𝑥 𝑥 = 𝑦)
3 dveeq1 1937 . . . . . . 7 (¬ ∀𝑥 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
43alimi 1385 . . . . . 6 (∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
54hbnaes 1652 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
6 19.21ht 1514 . . . . 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 1410 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦𝑥(𝑦 = 𝑧𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑)))
91, 8syl5rbbr 193 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑)))
10 sb6 1808 . 2 ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
11 sb6 1808 . . 3 ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧𝜑))
1211albii 1400 . 2 (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑))
139, 10, 123bitr4g 221 1 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wb 103  wal 1283  [wsb 1686
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469
This theorem depends on definitions:  df-bi 115  df-tru 1288  df-fal 1291  df-nf 1391  df-sb 1687
This theorem is referenced by: (None)
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