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Proof step 13 from proof shiftzero-not-neg (previous step)

i < 0, firstn(# fillsgn(n, int2bits(i)), mklist(false, i→n(j)) + fillsgn(n, int2bits(i))).first ⊦ j = 0, fillsgn(n, int2bits(i)) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ n ≠ 0 ∧ bits ≠ [] → (firstn(n, bits).first ↔ bits.first)
Proof step 15:
# fillsgn(n, int2bits(i)) ≠ 0 ∧ mklist(false, i→n(j)) + fillsgn(n, int2bits(i)) ≠ [], i < 0, (mklist(false, i→n(j)) + fillsgn(n, int2bits(i))).first ⊦ j = 0, fillsgn(n, int2bits(i)) = []

Rule argument

seq: ⊦ n ≠ 0 ∧ bits ≠ [] → (firstn(n, bits).first ↔ bits.first) subst: [bits, n] → [mklist(false, i→n(j)) + fillsgn(n, int2bits(i)), # fillsgn(n, int2bits(i))] rotate: F spec: list-del inst: bitlista name: firstn-first

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: mklist(false, i→n(j)) + fillsgn(n, int2bits(i)), # fillsgn(n, int2bits(i))