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Proof step 73 from proof int2bytes-rec-neg-05 (previous step)

i→n(1 + i * 8) % 8 = 0, 0 ≤ 2 ^ (i * 8) + j, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ i→n(1 + i * 8) / 8 = i→n(i) + 1, j = 0, bits2bytes(int2bits(j)) = i2b(~1) ' + fix(i→n(i) + 1, bits2bytes(int2bits(j))).rest

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 74:
⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1
Proof step 75:
0 ≤ i * 8, (i→n(i * 8) + 1) % 8 = 0, 0 ≤ 2 ^ (i * 8) + j, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ (i→n(i * 8) + 1) / 8 = i→n(i) + 1, j = 0, bits2bytes(int2bits(j)) = i2b(~1) ' + fix(i→n(i) + 1, bits2bytes(int2bits(j))).rest

Rule argument

seq: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1 subst: [i] → [i * 8] rotate: F spec: intnat inst: name: rax-04

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + j = j + i; Used terms: i * 8