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Proof step 21 from proof int2bytes-first-neg-01 (previous step)

int2bytes(i1 / 256) = (i2b(0) ' + bytes).butlast, int2bytes((i1 + 255) % 256).last = (i2b(0) ' + bytes).last, i1 < 0, i1 < ~383, ∀ bytes, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦ (i2b(0) ' + bytes).last ≠ i2b(0) ∧ ¬ (i2b(0) ' + bytes).last ∈ bytes, ¬ i2b(0) ∈ int2bytes(i1 / 256) ∧ i2b(0) ≠ (i2b(0) ' + bytes).last

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 22:
int2bytes(i1 / 256) = (i2b(0) ' + bytes).butlast, int2bytes((i1 + 255) % 256).last = (i2b(0) ' + bytes).last, i1 < 0, i1 < ~383, ∀ bytes, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦ bytes = [], (i2b(0) ' + bytes).last ≠ i2b(0) ∧ ¬ (i2b(0) ' + bytes).last ∈ bytes, ¬ i2b(0) ∈ int2bytes(i1 / 256) ∧ i2b(0) ≠ (i2b(0) ' + bytes).last

Rule argument

fma: bytes = []

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ bytes + [] = bytes;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by '.butlast = []; Used formulas: bytes = []