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

# bits < n ⊦ boolvar ∈ fillsgn(n, bits) ↔ boolvar ∈ bits, bits = []

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: ⊦ m < n → ¬ n < m;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ boolvar ∈ bits → boolvar ∈ bits0 + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ m < n → m ≤ n;% ⊦ boolvar ∈ mklist(boolvar0, n) → (boolvar ↔ boolvar0); (forward)% ⊦ boolvar ∈ mklist(boolvar, m) ↔ m ≠ 0;% ⊦ n < m → m - n ≠ 0;% ⊦ bits.first ∈ bits ↔ bits ≠ []; Used terms: bits, n