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Proof step 4 from proof int2bits-fillsgn (previous step)

∀ i0. mklist(i0 < 0, m) + int2bits(i0) = fillsgn(m + # int2bits(i0), int2bits(i0)) ⊦ ((¬ 0 ≤ i)' + mklist(¬ 0 ≤ i, m)) + int2bits(i) = fillsgn(# int2bits(i) + m + 1, int2bits(i))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits)
Proof step 6:
int2bits(i) ≠ [] ∧ # int2bits(i) < # int2bits(i) + m + 1, ∀ i0. mklist(i0 < 0, m) + int2bits(i0) = fillsgn(m + # int2bits(i0), int2bits(i0)) ⊦ ((¬ 0 ≤ i)' + mklist(¬ 0 ≤ i, m)) + int2bits(i) = int2bits(i).first ' + fillsgn((# int2bits(i) + m + 1)-1, int2bits(i))

Rule argument

seq: ⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits) subst: [bits, n] → [int2bits(i), # int2bits(i) + m + 1] rotate: F spec: bitopsaux inst: name: fill-rec

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ m + n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: int2bits(i), # int2bits(i) + m + 1