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Theorem sbal2 1998
 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 1700 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑦 ¬ ∀𝑥 𝑥 = 𝑦)
2 dveeq1 1995 . . . . . . 7 (¬ ∀𝑥 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
32alimi 1432 . . . . . 6 (∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
43hbnaes 1702 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
5 19.21ht 1561 . . . . 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 1457 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦𝑥(𝑦 = 𝑧𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑)))
8 alcom 1455 . . 3 (∀𝑦𝑥(𝑦 = 𝑧𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑))
97, 8bitr3di 194 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑧𝜑)))
10 sb6 1859 . 2 ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
11 sb6 1859 . . 3 ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧𝜑))
1211albii 1447 . 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 1330  [wsb 1736 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 604  ax-in2 605  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516 This theorem depends on definitions:  df-bi 116  df-tru 1335  df-fal 1338  df-nf 1438  df-sb 1737 This theorem is referenced by: (None)
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