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Proof step 6 from proof int2bytes-len-pos-01 (previous step)

# int2bits(j) % 8 = 0, # int2bits(j) % 8 = 0, 0 ≤ i, j < 2 ^ (8 * (i + 1)), 2 ^ (8 * (i + 1) - 1) ≤ j ⊦ # int2bits(j) / 8 = i→n(i + 2)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1) ⊦ # int2bits(j) = i→n(i + 2)
Proof step 8:
¬(0 ≤ 8 * (i + 1) ∧ 2 ^ (8 * (i + 1)) ≤ j ∧ j < 2 ^ (8 * (i + 1) + 1)), # int2bits(j) % 8 = 0, # int2bits(j) % 8 = 0, 0 ≤ i, j < 2 ^ (8 * (i + 1)), 2 ^ (8 * (i + 1) - 1) ≤ j ⊦ # int2bits(j) / 8 = i→n(i + 2)

Rule argument

seq: 0 ≤ i, 2 ^ i ≤ j, j < 2 ^ (i + 1) ⊦ # int2bits(j) = i→n(i + 2) subst: [i, j] → [8 * (i + 1), j] rotate: F spec: int2bits inst: name: int2bits-len-pos

Additional information

Used simplifier rules: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false); Used terms: 8 * (i + 1), j