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Proof step 9 from proof add-eq2pot (previous step)

2 ^ (i1 + i0) = 2 ^ i0 + j, 2 ^ i1 = i2 + 1, 0 ≤ i1 + i0, 0 ≤ j, 0 ≤ i0 ⊦ j = 2 ^ i0 * i2 ∧ 0 ≤ i1, 0 ≤ i1 ∧ 0 ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
2 ^ (i1 + i0) = 2 ^ i0 + j, i1 < 0, 0 ≤ i1 + i0, 0 ≤ j, 0 ≤ i0 ⊦ j = 2 ^ i0 * ~1 ∧ 0 ≤ i1

Rule argument

Additional information

Used simplifier rules: two-minus: ⊦ i < 0 → 2 ^ i = 0;% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)