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Proof step 5 from proof fill-empty (previous step)

fillsgn(n, bits) = [], # bits < n ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ # bits ≤ n ∧ bits ≠ [] → fillsgn(n, bits) = mklist(bits.first, n - # bits) + bits

Rule argument

name: fill-fill rotate: F subst: [bits, n] → [bits, n]

Additional information

Used simplifier rules: fillsgn-empty: ⊦ fillsgn(n, []) = mklist(false, n);% ⊦ m < n → n ≠ 0;% ⊦ m < n → ¬ n < m;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ n < m → m - n ≠ 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: bits, n