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Proof step 13 from proof shiftzero-by-one (previous step)

i < ~1, i < 0 ⊦ i = ~1, bits2int(firstn(# fillsgn(m + 1, int2bits((i - 1) / 2) + (i % 2 ≠ 0)'), false ' + fillsgn(m + 1, int2bits((i - 1) / 2) + (i % 2 ≠ 0)'))) = bits2uint(fillsgn(m, int2bits((i - 1) / 2)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ 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), m + 1] rotate: F spec: bitopsaux inst: name: fill-last

Additional information

Used simplifier rules: ⊦ (m + 1)-1 = m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ firstn(n + 1, boolvar ' + bits) = boolvar + firstn(n, bits);% ⊦ firstn(# bits, bits + bits0) = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bits2int(false ' + bits) = bits2uint(bits);% ⊦ int2bits(j) ≠ [];% ⊦ m + 1 ≠ 0; Used terms: i % 2 ≠ 0, int2bits((i - 1) / 2), m + 1