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Proof step 58 from proof no-leading-zero-01 (previous step)

bytes = [], int2bytes(i / 256) = i2b(0) ' + (by ' + bytes).butlast, int2bytes(i % 256).last = (by ' + bytes).last, 0 < i, 0 ≤ b2i(by), 255 < i, i2b(0) ∈ int2bytes(i / 256), ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + by ' + bytes ∧ 0 ≤ b2i(by) ∧ i2b(0) ∈ int2bytes(i0) → false ⊦ (by ' + bytes).last ≠ i2b(0) ∧ (by ' + bytes).last ≠ by ∧ ¬ (by ' + bytes).last ∈ bytes

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: int2bytes-zero: ⊦ int2bytes(i) = i2b(0) ' ↔ i = 0;%i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j;% ⊦ by '.last = by;%0 < j ⊦ 0 < i ∧ i < j → i % (* j) = i;% ⊦ 0 < i ∧ 0 < j → (i / j = 0 ↔ i < j);% ⊦ by '.butlast = [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + [] = bytes;