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Proof step 18 from proof bint-is-cast (previous step)

32 < # int2bits(i) ⊦ bi2i(i2bi(cast(n→i(32), i))) = cast(32, i), i ∈bint

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
32 < # int2bits(i) ⊦ bi2i(i2bi(cast(32, i))) = cast(32, i), i ∈bint

Rule argument

Additional information

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