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Proof step 12 from proof int2bits-len-eq (previous step)

j = 2 ^ n→i(# int2bits(j)), ¬ j ≤ 0 → 0 ≤ i ∧ 2 ^ i ≤ j ∧ j < 2 ^ (i + 1) ⊦ j < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
j = 2 ^ n→i(# int2bits(j)), 0 ≤ i, 0 < j, j < 2 ^ i * 2, 2 ^ i ≤ j ⊦ 0 = j

Rule argument

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';% ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ # boolvar ' = 0 +1;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;