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Proof step 15 from proof dec-int-01 (previous step)

2 * i→n(i) = m + 1, m = m0 * 2 + n0, i < 0, n0 < 2 ⊦ n→i((m + 1) / 2) + i = 0, 2 = 0, i = 0, n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m0 + 1 = i→n(i), i < 0 ⊦ ~1 = n→i(m0) + i, i = 0

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)% ⊦ m * n + n = m1 * n ↔ m + 1 = m1 ∨ n = 0; (s)% ⊦ m ≠ 0 → (m < 2 ↔ m = 1); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ n→i(m + n) = n→i(m) + n→i(n); (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% ⊦ (2 * m + n) / 2 = m + n / 2; (s)