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Proof step 21 from proof and-com (previous step)

∀ bits1, bits. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ bitsand(bits, bits1) = bitsand(bits1, bits) ⊦ bitsand(boolvar ' + bits2, bits0) = bitsand(bits0, boolvar ' + bits2), bits0 = [], bits0 = true ', bits0 = false ', boolvar ' + bits2 = true ', boolvar ' + bits2 = false ', bits0 ≠ bits0.first ' ∧ bits2 ≠ []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
bits0 = boolvar0 + bits3, ∀ bits1, bits. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ bitsand(bits, bits1) = bitsand(bits1, bits) ⊦ bitsand(boolvar ' + bits2, bits0) = bitsand(bits0, boolvar ' + bits2), bits0 = [], bits0 = true ', bits0 = false ', boolvar ' + bits2 = true ', boolvar ' + bits2 = false ', bits0 ≠ boolvar0 ' ∧ bits2 ≠ []

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits0, boolvar0, bits3] 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, boolvar0, bits3