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Proof step 6 from proof int2bytes-last-01 (previous step)

¬ i < 128 ∧ ¬ 256 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (i2b(0) ' + i2b(i - 256) ').last = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ i, 0 < i, 128 ≤ i, i < 256 ⊦ i2b(i - 256) = i2b(i)

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ (bytes + by ').last = by;