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Proof step 36 from proof int2bytes-first-neg (previous step)

int2bytes(i) = by ' + bytes0, i ≤ 0, j ≤ 0, 0 ≤ b2i(by), i * 256 + j < ~383, abs(j) < 256, by ∈ int2bytes(i), 0 ≤ (i + 1) * 255 + j, i * 256 + j < 0, i * 256 < (i + 1) * 255, ~1 ≤ i, ∀ bytes, by, i0. i→n(i0) < i→n((i * 256 + j) + 255) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = ~1

Additional information

Used simplifier rules: int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';%int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);% ⊦ i - j < i0 ↔ i < j + i0;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ i * j = j * i;% ⊦ by ' = by0 ' + bytes ↔ by = by0 ∧ bytes = [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ i + 0 = i;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ by ∈ by0 ' ↔ by = by0;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;%i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) Used formulas: i = ~1