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Proof step 45 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(mklist(true, n), bits0)) = fillsgn(n, bits0), bits0 = [], n = 0, ¬ n < # bits0 ∧ bits0.first

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 46:
bits0 = boolvar + bits2, ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(mklist(true, n), bits0)) = fillsgn(n, bits0), bits0 = [], n = 0, ¬ n < # bits0 ∧ boolvar

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits0, boolvar, bits2] 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: bits0, boolvar, bits2