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Proof step 2 from proof le-2pot-succ (previous step)

0 < i, 0 < i0, 0 < j ⊦ 2 ^ i ≤ j * 2 ^ i0 + 1 ↔ 2 ^ i ≤ j * 2 ^ i0

User applied rule case distinction

And obtained the following new goals

Proof step 3:
2 ^ i ≤ j * 2 ^ i0 + 1, 0 < i, 0 < i0, 0 < j ⊦ 2 ^ i ≤ j * 2 ^ i0

Rule argument

pos: right 1

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2