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Proof step 9 from proof shiftzero-not-neg (previous step)

sublist(0, # fillsgn(n, int2bits(i)), mklist(false, i→n(j)) + fillsgn(n, int2bits(i))) ≠ [], sublist(0, # fillsgn(n, int2bits(i)), mklist(false, i→n(j)) + fillsgn(n, int2bits(i))).first, i < 0 ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
sublist(0, # fillsgn(n, int2bits(i)), mklist(false, i→n(j)) + fillsgn(n, int2bits(i))).first, i < 0 ⊦ j = 0, fillsgn(n, int2bits(i)) = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ sublist(0, n, bits) = [] ↔ n = 0;