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Proof step 4 from proof mod-two-pred-01 (previous step)

2 * i - 1 = i0 * 2 + j0 ∧ j0 ≤ 0 ∧ abs(j0) < 2 ∧ i0 ≤ 0 ∧ 0 < 2 ∧ 2 * i - 1 < 0, i < 0 ⊦ j0 = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 * i = (i0 * 2 + j0) + 1, abs(j0) < 2, i < 0, j0 ≤ 0, i0 ≤ 0 ⊦ j0 = ~1

Rule argument

Additional information

Used simplifier rules: nummult-ls: ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; 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%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs