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

# bits < n ⊦ fillsgn(n, bits) = bits.first ' + fillsgn(n - 1, bits), bits = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n = m + 1 ∧ n ≠ 0, # bits < n ⊦ fillsgn(n, bits) = bits.first ' + fillsgn(m, bits), bits = []

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: ⊦ bits = bits0 + bits ↔ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ n < 0;%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m