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Proof step 22 from proof band-min-02 (previous step)

int2bits(i).first ∧ # int2bits(i) ≤ # int2bits(j) ∧ int2bits(i) ≠ [], j < 0, i < 0, i0 + 2 ^ (n→i(# int2bits(j)) - 1) ≤ 0, i0 + 2 ^ (n→i(# int2bits(i)) - 1) ≤ 0, bits2int(bitsand(int2bits(j), int2bits(i))) < i0, # int2bits(i) < # int2bits(j), ¬ (# int2bits(j) - 1) + 1 < # int2bits(j) → 0 ≤ 2 ^ n→i(# int2bits(j) - 1) + bits2int(bitsand(int2bits(j), int2bits(i))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
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 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ n + (m - n) < m;% ⊦ m < n → m ≤ n;% ⊦ j < 0 → int2bits(j).first;% ⊦ int2bits(j) ≠ [];