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Proof step 7 from proof shiftright-not-pos-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i < 0 ⊦ sublist(0, # int2bits(i), mklist(true, i→n(j)) + int2bits(i)).first

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ sublist(0, n, bits) = [] ↔ n = 0;