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Proof step 31 from proof int2bytes-last-01 (previous step)

~256 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i, ~256 = i ∨ ~128 ≤ i ⊦ ~256 = i, int2bytes(i % 256).last = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
256 ≤ i, 0 < i, 0 ≤ i ⊦ ~256 = i, int2bytes(i % 256).last = i2b(i)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);%i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j;% ⊦ i ≠ j → (i ≤ j ↔ i < j);