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Proof step 4 from proof bint-fillsgn-len (previous step)

¬ # int2bits(i) ≤ 32, ~2147483648 ≤ i, i ≤ 2147483647 ⊦ # fillsgn(32, int2bits(i)) = 32

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
32 < # int2bits(i), i ≤ 2147483647, ~2147483648 ≤ i ⊦ # fillsgn(32, int2bits(i)) = 32

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;