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Proof step 16 from proof and-ones (previous step)

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

User applied rule insert elim lemma

And closed the goal!

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar ' subst: [bits, boolvar, bits0] → [bits2, boolvar, bits] rotate: T spec: list-last inst: bitlista name: last-elim

Additional information

Used simplifier rules: ⊦ bits + boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;%bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar '; (elim) Used terms: bits2, boolvar, bits