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Proof step 19 from proof and-onesin (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
bits = boolvar0 + bits2, ∀ bits1, bits0, boolvar. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ boolvar ∈ bitsand(bits, bits0) ↔ boolvar ∈ bits0, bits0 = [], bits = [], bits0 = true ', bits0 = false ', bits = true ', bits = false ', false ∈ bits, bits ≠ boolvar0 ' ∧ bits0 ≠ bits0.first '

Rule argument

seq: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0) subst: [bits, boolvar, bits0] → [bits, boolvar0, bits2] rotate: T spec: list-data inst: bitlista name: elim-1

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0); (elim) Used terms: bits, boolvar0, bits2