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Proof step 8 from proof fill-lastn-03 (previous step)

m < n, # bits < n, m < # bits, m + 1 < n ⊦ bits = [], lastn(m, fillsgn(n, bits)) = lastn(m, bits)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ # bits ≤ n ∧ bits ≠ [] → fillsgn(n, bits) = mklist(bits.first, n - # bits) + bits
Proof step 10:
# bits ≤ n ∧ bits ≠ [], m < n, # bits < n, m < # bits, m + 1 < n ⊦ bits = [], lastn(m, mklist(bits.first, n - # bits) + bits) = lastn(m, 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; Used terms: bits, n