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Proof step 35 from proof int2bytes-first-neg-01 (previous step)

int2bytes(i) = i2b(0) ' + bytes0, j < 0, ~256 < j, i2b(0) ∈ int2bytes(i), 0 ≤ 255 * (i + 1) + j, i * 256 + j < 0, i * 256 < 255 * (i + 1), ~1 ≤ i, i ≤ 0, i * 256 + j < ~383, ∀ bytes, i0. i→n(i0) < i→n((i * 256 + j) + 255) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦ j = 0

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) ';%~128 ≤ i, i ≤ 127, ~128 ≤ j, j ≤ 127, i ≠ j ⊦ i2b((* i)) ≠ i2b((* j));% ⊦ 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);% ⊦ by ' = by0 ' + bytes ↔ by = by0 ∧ bytes = [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ i + 0 = i;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ by ∈ by0 ' ↔ by = by0;%i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) Used formulas: i = ~1