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Proof step 22 from proof fillsgn-butlast (previous step)

b2i(bytes.butlast.first) < 0, # bytes ≤ n, 0 ≤ b2i(bytes.first) ⊦ bytes = bytes.first ', fillsgn(n, bytes.butlast) = mklist(i2b(0), (n - # bytes) + 1) + bytes.butlast, bytes = [], # bytes = 1

User applied rule insert elim lemma

And closed the goal!

Rule argument

seq: bytes ≠ [] ⊦ by = bytes.last ∧ bytes0 = bytes.butlast ↔ bytes = bytes0 + by ' subst: [bytes, by, bytes0] → [bytes, by, bytes0] rotate: T spec: list-last inst: bytelist name: last-elim

Additional information

Used simplifier rules: ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ # by ' = 0 +1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + by ' ≠ [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ bytes ≠ [] → (bytes + bytes0).first = bytes.first;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);%bytes ≠ [] ⊦ by = bytes.last ∧ bytes0 = bytes.butlast ↔ bytes = bytes0 + by '; (elim) Used terms: bytes, by, bytes0