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Proof step 52 from proof int2bytes-len-neg-01 (previous step)

# int2bits(j) = 1, j < 0, j + 2 ^ (8 * (i + 1) - 1) < 0, 0 ≤ i ⊦ j + 2 ^ (8 * (i + 1)) = 0, ~1 = i

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ j < 0 ∧ j ≠ ~1 → int2bits(j) = int2bits((j - 1) / 2) + (~1 = j % 2)' subst: [j] → [j] rotate: F spec: int2bits inst: name: int2bits-rec-neg

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ 2 ^ i = 1 ↔ i = 0;% ⊦ i * j = 0 ↔ i = 0 ∨ j = 0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ 2 ^ i < 1 ↔ i < 0;% ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false); Used terms: j