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Proof step 8 from proof i2n-divtwo (previous step)

~(2 * i) = n→i(m0 * 2 + n0), i < 0, n0 < 2, 2 * i < 0 ⊦ m0 = i→n(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
2 * (n→i(m0) + i) + n→i(n0) = 0, n0 < 2, i < 0 ⊦ m0 = i→n(i)

Rule argument

Additional information

Used simplifier rules: n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%n2i-mult: ⊦ n→i(m * n) = n→i(m) * n→i(n);% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * j + i * j0 = i * (j + j0); (s)% ⊦ ~ i = j ↔ i + j = 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult