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Proof step 34 from proof int2bytes-last-01 (previous step)

¬(0 ≤ i % 256 ∧ i % 256 ≤ 127), 256 ≤ i, 0 < i, 0 ≤ i ⊦ ~256 = i, int2bytes(i % 256).last = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
0 < i % 256, 127 < i % 256, 0 ≤ i, 0 < i, 256 ≤ i ⊦ int2bytes(i % 256).last = i2b(i), ~256 = i

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j;