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Proof step 15 from proof cast-size-fits (previous step)

0 < j, 0 ≤ i0, j < 2 ^ (n→i(i→n(i0) + 1) - 1) ⊦ cast(i0 + 1, j) = j, ¬ 2 ^ i0 ≤ j ∧ ¬ 2 ^ i0 + j < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
0 < j, j < 2 ^ i0, 0 ≤ i0, 2 ^ i0 + j < 0, 0 < i0 ⊦ cast(i0 + 1, j) = j

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < 2 ^ i ↔ 0 < i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 ≤ i → n→i(i→n(i)) = i;% ⊦ n→i(m + n) = n→i(m) + n→i(n);