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Proof step 4 from proof cast-length (previous step)

0 < i, i→n(i) < # int2bits(bits2int(lastn(i→n(i), fillsgn(i→n(i), int2bits(j))))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ sgn(bits) = [] ↔ bits = [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)%bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits);% ⊦ lastn(n, bits) = [] ↔ n = 0;% ⊦ i→n(i) = 0 ↔ i = 0;