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Proof step 1 from proof bint-fillsgn-len (conclusion)

⊦ i ∈bint → # fillsgn(32, int2bits(i)) = 32

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
~2147483648 ≤ i, i ≤ 2147483647 ⊦ # fillsgn(32, int2bits(i)) = 32

Rule argument

Additional information

Used simplifier rules: inbint: ⊦ i ∈bint ↔ ~2147483648 ≤ i ∧ i ≤ 2147483647;% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)