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Proof step 14 from proof shiftzero-by-one-01 (previous step)

i < ~1, i < 0 ⊦ i = ~1, fillsgn(n - 1, int2bits((i - 1) / 2)) = [], n = 0, 2 ^ n→i(# fillsgn(n - 1, int2bits((i - 1) / 2))) = 2 ^ n→i(# fillsgn(n, int2bits((i - 1) / 2) + (i % 2 ≠ 0)') - 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ n ≠ 0 ∧ bits ≠ [] → fillsgn(n, bits + boolvar ') = fillsgn(n -1, bits) + boolvar

Rule argument

seq: ⊦ n ≠ 0 ∧ bits ≠ [] → fillsgn(n, bits + boolvar ') = fillsgn(n -1, bits) + boolvar subst: [boolvar, bits, n] → [i % 2 ≠ 0, int2bits((i - 1) / 2), n] rotate: F spec: bitopsaux inst: name: fill-last

Additional information

Used simplifier rules: ⊦ n -1 = n - 1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ 0 + m = m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ int2bits(j) ≠ []; Used terms: i % 2 ≠ 0, int2bits((i - 1) / 2), n