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Proof step 18 from proof int2bytes-rec-case (previous step)

i % 256 = 0, i < ~256, i < 0 ⊦ i = ~256, int2bytes(i) = int2bytes(i / 256) + i2b(0) '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ i < ~256 ∧ i % 256 = 0 → int2bytes(i) = int2bytes(i / 256) + int2bytes(i % 256).last '

Rule argument

name: int2bytes-rec-neg-256 rotate: F subst: [i] → [i]

Additional information

Used simplifier rules: int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';% ⊦ i < j → ¬ j < i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;% ⊦ by ' = by0 ' ↔ by = by0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by '.last = by;%~128 ≤ i, i ≤ 127, ~128 ≤ j, j ≤ 127 ⊦ i2b((* i)) = i2b((* j)) ↔ i = j; Used terms: i