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Proof step 21 from proof fill-last (previous step)

# bits ≤ m0, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ bits = [], fillsgn(2 + m0, bits + boolvar ') = bits.first ' + fillsgn(m0, bits) + boolvar '

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
bits = boolvar0 + bits0, # bits ≤ m0, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ bits = [], fillsgn(2 + m0, bits + boolvar ') = boolvar0 ' + fillsgn(m0, bits) + boolvar '

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits, boolvar0, bits0] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar0, bits0