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Proof step 11 from proof mod-two-ge-01 (previous step)

0 ≤ j, 0 < i0, 0 < i0 + j ⊦ 2 ^ j * 2 ^ i0 = 2 ^ j * ((2 ^ j * 2 ^ i0 - 1) / 2 ^ j + 1), 1 < 2 ^ j ∧ 0 < 2 ^ i0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [2 ^ j * 2 ^ i0, 1, i] flags:

Additional information

Used simplifier rules: pot2-pos-01: ⊦ 1 < 2 ^ i ↔ 0 < i;%pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%one: ⊦ i / 1 = i; from specification int-div%add-in: ⊦ i + j = i + i0 ↔ j = i0; 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% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: 2 ^ j * 2 ^ i0, 1, i