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Proof step 5 from proof int2bits-len-range (previous step)

i→n(i + 1) < # int2bits(j), 0 ≤ 2 ^ i + j, j < 2 ^ i, j ≠ 0 → 0 ≤ i0 ∧ 2 ^ i0 ≤ abs(j) ∧ abs(j) < 2 ^ (i0 + 1) ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 6:
i→n(i + 1) < # int2bits(j), 0 ≤ 2 ^ i + j, j < 2 ^ i, j ≠ 0 → 0 ≤ i0 ∧ 2 ^ i0 ≤ abs(j) ∧ abs(j) < 2 ^ (i0 + 1) ⊦ j = 0

Rule argument

fma: j = 0

Additional information

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