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Proof step 11 from proof uint2bits-add-high-01 (previous step)

# int2bits(j0) ≤ i→n(i) + 1, 1 < j, 0 < j, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 ≤ j0 + 2 ^ i, 0 ≤ i ⊦ fillsgn(i→n(i) + 1, sgn(int2bits(j0).rest)).rest = fillsgn(i→n(i), int2bits(j0).rest)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ bits ≠ [] ∧ # bits ≤ n + 1 → fillsgn(n + 1, sgn(bits.rest)).rest = fillsgn(n, bits.rest)

Rule argument

seq: ⊦ bits ≠ [] ∧ # bits ≤ n + 1 → fillsgn(n + 1, sgn(bits.rest)).rest = fillsgn(n, bits.rest) subst: [bits, n] → [int2bits(j0), i→n(i)] rotate: F spec: bitopsaux inst: name: fill-sgn-03

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m ≤ n ∧ n < n0 → m < n0; (forward)% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n < m + n0 ↔ n < n0; Used terms: int2bits(j0), i→n(i)