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

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

User applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ boolvar + bits = boolvar ' + bits;% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];%⊦ ¬ boolvar ∈ boolvar0 ' + bits ↔ ¬(boolvar ↔ boolvar0) ∧ ¬ boolvar ∈ bits;%⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0); (elim) Used terms: bits0, boolvar0, bits