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Proof step 2 from proof fill-one-02 (previous step)

0 < i ⊦ fillsgn(i→n(i), boolvar ') = mklist(boolvar, i→n(i))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ # bits ≤ n ∧ bits ≠ [] → fillsgn(n, bits) = mklist(bits.first, n - # bits) + bits
Proof step 4:
# boolvar ' ≤ i→n(i) ∧ boolvar ' ≠ [], 0 < i ⊦ mklist(boolvar '.first, i→n(i) - # boolvar ') + boolvar ' = mklist(boolvar, i→n(i))

Rule argument

name: fill-fill rotate: F subst: [bits, n] → [boolvar ', i→n(i)]

Additional information

Used simplifier rules: ⊦ # boolvar ' = 0 +1;% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;%0 ≤ i ⊦ (* m) ≤ i→n(i) ↔ n→i(m) ≤ i;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ boolvar ' ≠ []; Used terms: boolvar ', i→n(i)