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Proof step 7 from proof modex-odd-neg (previous step)

i = i0 * 2 + j0 ∧ j0 ≤ 0 ∧ abs(j0) < 2 ∧ i0 ≤ 0 ∧ 0 < 2, i < 0 ⊦ j0 = ~1 ↔ (∃ j. 2 * j = i + 1 ∧ j ≤ 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
2 * j = (i0 * 2 + j0) + 1, abs(j0) < 2, j0 ≤ 0, i0 ≤ 0, i0 * 2 + j0 < 0, j ≤ 0 ⊦ j0 = ~1

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%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%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%add-zero: ⊦ i + 0 = i; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2