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Proof step 2 from proof fill-lastn-05 (previous step)

# bits0 ≤ m ⊦ lastn(m, fillsgn(m + 1, bits0).rest) = fillsgn(m, bits0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits)
Proof step 4:
bits0 ≠ [] ∧ # bits0 < m + 1, # bits0 ≤ m ⊦ lastn(m, (bits0.first ' + fillsgn((m + 1)-1, bits0)).rest) = fillsgn(m, bits0)

Rule argument

name: fill-rec rotate: F subst: [bits, n] → [bits0, m + 1]

Additional information

Used simplifier rules: fillsgn-empty: ⊦ fillsgn(n, []) = mklist(false, n);% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m ≤ n ∧ n < n0 → m < n0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ # bits = 0 ↔ bits = [];% ⊦ ¬ n < n;% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);% ⊦ mklist(boolvar, 1) = boolvar ';% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ lastn(n, mklist(boolvar, n)) = mklist(boolvar, n);% ⊦ mklist(boolvar, m) = mklist(boolvar, n) ↔ m = n; Used terms: bits0, m + 1