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Proof step 7 from proof minus-odd (previous step)

0 < 2 * j0 + 1, 0 ≤ j0 ⊦ i ^ (2 * j0 + 1) + ~ i ^ (2 * j0 + 1)-1 * ~ i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: minus-square: ⊦ 0 ≤ j → ~ i ^ (2 * j) = i ^ (2 * j);%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%pred: ⊦ (i + 1)-1 = i; from specification int-basic2