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Proof step 10 from proof bint-fillsgn-len (previous step)

¬ 0 < i, 32 < # int2bits(i), i ≤ 2147483647, ~1 ≤ i ⊦ # fillsgn(32, int2bits(i)) = 32

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
32 < # int2bits(i), ~1 ≤ i, i ≤ 0 ⊦ # fillsgn(32, int2bits(i)) = 32

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0);% ⊦ ¬ i < j ↔ j ≤ i;