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Proof step 8 from proof shiftzero-by-one-01 (previous step)

i < ~1, i < 0 ⊦ n = 0, bits2int(fillsgn(n - 1, int2bits((i - 1) / 2))) + 2 ^ n→i(# fillsgn(n - 1, int2bits((i - 1) / 2))) = (i - 1) / 2 + 2 ^ n→i(# fillsgn(n, int2bits(i)) - 1), fillsgn(n - 1, int2bits((i - 1) / 2)) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ bits ≠ [] → bits2int(fillsgn(n, bits)) = bits2int(bits)
Proof step 10:
int2bits((i - 1) / 2) ≠ [], i < ~1, i < 0 ⊦ n = 0, bits2int(int2bits((i - 1) / 2)) + 2 ^ n→i(# fillsgn(n - 1, int2bits((i - 1) / 2))) = (i - 1) / 2 + 2 ^ n→i(# fillsgn(n, int2bits(i)) - 1), fillsgn(n - 1, int2bits((i - 1) / 2)) = []

Rule argument

seq: ⊦ bits ≠ [] → bits2int(fillsgn(n, bits)) = bits2int(bits) subst: [bits, n] → [int2bits((i - 1) / 2), n - 1] rotate: F spec: int2bits inst: name: bits2int-fill

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ []; Used terms: int2bits((i - 1) / 2), n - 1