[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 12 from proof pot-square-est-div6 (previous step)

0 ≤ i / 6 ∧ 0 ≤ 2, 0 ≤ 2 ∧ 0 ≤ 2 ∧ 0 ≤ i / 6, (2 ^ (i / 6)) ^ 2 ≤ (i + 1) ^ 2, 36 ≤ i, 0 ≤ i, 0 < i ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i ^ j ≤ i0 ^ j ↔ i ≤ i0)
Proof step 14:
0 ≤ 2 ^ (i / 6) ∧ 0 ≤ i + 1 ∧ 0 < 2, 0 ≤ i / 6 ∧ 0 ≤ 2, 0 ≤ 2 ∧ 0 ≤ 2 ∧ 0 ≤ i / 6, 2 ^ (i / 6) ≤ i + 1, 36 ≤ i, 0 ≤ i, 0 < i ⊦

Rule argument

spec: inst: name: le-pos-01 seq: ⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i ^ j ≤ i0 ^ j ↔ i ≤ i0) subst: [i0, j, i] → [i + 1, 2, 2 ^ (i / 6)] flags: rewrite, current

Additional information

Used simplifier rules: two-lsz: ⊦ ¬ 2 ^ i < 0;%zero-03: ⊦ 0 < i ∧ 0 < j → 0 ≤ i / j; from specification int-div%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs Used terms: i + 1, 2, 2 ^ (i / 6)