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Proof step 24 from proof band-min-02 (previous step)

2 ^ n→i(# int2bits(j) - 1) + bits2int(bitsand(int2bits(j), int2bits(i))) < 2 ^ n→i(# int2bits(j) - 1) + i0, 0 ≤ 2 ^ n→i(# int2bits(j) - 1) + bits2int(bitsand(int2bits(j), int2bits(i))), # int2bits(i) < # int2bits(j), bits2int(bitsand(int2bits(j), int2bits(i))) < i0, i0 + 2 ^ (n→i(# int2bits(i)) - 1) ≤ 0, i0 + 2 ^ (n→i(# int2bits(j)) - 1) ≤ 0, i < 0, j < 0 ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: 0 < 2 ^ n→i(# int2bits(j) - 1) + i0

Additional information

Used simplifier rules: ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i + j ≤ i0 + j ↔ i ≤ i0;% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 0 < n ↔ n ≠ 0;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;%0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n;% ⊦ m ≤ m - 1 → m = 0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) Used formulas: 0 < 2 ^ n→i(# int2bits(j) - 1) + i0