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Proof step 11 from proof uint2bits-len-add (previous step)

0 ≤ i, 0 ≤ j0, 0 < j, j0 < 2 ^ i ⊦ # fillsgn(i→n(i) + 1, uint2bits(j0)) - 1 = i→n(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ # bits ≤ n → # fillsgn(n, bits) = n subst: [bits, n] → [uint2bits(j0), i→n(i) + 1] rotate: F spec: bitopsaux inst: name: fill-length

Additional information

Used simplifier rules: ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ 0 + m = m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ i→n(i) = i→n(j) ↔ abs(i) = abs(j); Used terms: uint2bits(j0), i→n(i) + 1