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Proof step 27 from proof int2bytes-leading-zeros (previous step)

255 < i, int2bytes(i / 256) + int2bytes(i % 256).last ' = i2b(0) ' + i2b(0) ' + bytes, 0 ≤ i, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
int2bytes(i / 256) + int2bytes(i % 256).last ' = i2b(0) ' + i2b(0) ' + bytes, 0 < i, 0 ≤ i, 255 < i, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦ int2bytes(i % 256).last ≠ i2b(0) ∧ ¬ int2bytes(i % 256).last ∈ bytes, ¬ i2b(0) ∈ int2bytes(i / 256) ∧ i2b(0) ≠ int2bytes(i % 256).last

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ by ≠ by0 → (by ∈ by0 ' + bytes ↔ by ∈ bytes);% ⊦ by ∈ by0 ' ↔ by = by0;% ⊦ by ∈ bytes + bytes0 ↔ ¬ ¬(by ∈ bytes ∨ by ∈ bytes0);% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + by ' = bytes0 → by ∈ bytes0; (forward)