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Proof step 2 from proof fillsgn-rec (previous step)

# bytes < n ⊦ fillsgn(n, bytes) = (¬ 0 ≤ b2i(bytes.first) ⊃ i2b(~1);i2b(0)) ' + fillsgn(n - 1, bytes), bytes = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n = m + 1 ∧ n ≠ 0, # bytes < n ⊦ fillsgn(n, bytes) = (¬ 0 ≤ b2i(bytes.first) ⊃ i2b(~1);i2b(0)) ' + fillsgn(m, bytes), bytes = []

Rule argument

seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m] rotate: T spec: nat inst: name: elim-pred-c

Additional information

Used simplifier rules: ⊦ bytes = bytes0 + bytes ↔ bytes0 = [];% ⊦ by ' ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ ¬ n < 0;%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m