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Proof step 7 from proof bint-fillsgn-len (previous step)

¬ i < ~1, 32 < # int2bits(i), i ≤ 2147483647, ~2147483648 ≤ i ⊦ # fillsgn(32, int2bits(i)) = 32

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
32 < # int2bits(i), i ≤ 2147483647, ~1 ≤ i ⊦ # fillsgn(32, int2bits(i)) = 32

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;