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Proof step 16 from proof band-min-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
j < 0, i < 0, i0 + 2 ^ (n→i(# int2bits(j)) - 1) ≤ 0, i0 + 2 ^ (n→i(# int2bits(i)) - 1) ≤ 0, bits2int(bitsand(int2bits(i), int2bits(j))) < i0, # int2bits(i) < # int2bits(j), ¬ (# int2bits(j) - 1) + 1 < # bitsand(int2bits(i), int2bits(j)) → 0 ≤ 2 ^ n→i(# int2bits(j) - 1) + bits2int(bitsand(int2bits(i), int2bits(j))) ⊦

Rule argument

Additional information

Used simplifier rules: and-notempty-09: ⊦ bitsand(int2bits(i), int2bits(i0)) ≠ [];% ⊦ int2bits(j) ≠ [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = [];