[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 8 from proof exin2pot (previous step)

0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1)) ⊦ j = 1, j = 0, ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)

User applied rule apply induction

And obtained the following new goals

Proof step 9:
0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1)) ⊦ 0 < j / 2, j = 1, j = 0, ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)
Proof step 12:
∃ i. 0 ≤ i ∧ 2 ^ i ≤ j / 2 ∧ j / 2 < 2 ^ (i + 1), 0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (∃ i. 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1)) ⊦ j = 1, j = 0, ∃ i. 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1)

Rule argument

subst: [j0] → [j / 2]

Additional information

Used simplifier rules: labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%abs: j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j); from specification int-div%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%div2-nle: ⊦ 0 < i → ¬ i ≤ i / 2; from specification int-div Used terms: j / 2