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Proof step 5 from proof fill-fill-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ # bits ≤ n → # fillsgn(n, bits) = n

Rule argument

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

Additional information

Used simplifier rules: fill-nls: ⊦ ¬ # fillsgn(n, bits) < # bits;% ⊦ ¬ n < n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ # bits = 0 ↔ bits = [];% ⊦ m +1 = m + 1; Used terms: bits, n