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Proof step 18 from proof bitsand-first-samesize (previous step)

# bits1 = 1, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < 1 → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bits1 = [], bitsand(boolvar ' + bits1.butlast, boolvar0 ') = (boolvar ∧ boolvar0)'

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] → [bits1, boolvar1, bits] rotate: T spec: list-last inst: bitlista name: last-elim

Additional information

Used simplifier rules: and-one: ⊦ bitsand(boolvar ', boolvar0 ') = (boolvar ∧ boolvar0)';% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ bits + boolvar ' ≠ [];% ⊦ bits + [] = bits;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;%bits ≠ [] ⊦ (boolvar ↔ bits.last) ∧ bits0 = bits.butlast ↔ bits = bits0 + boolvar '; (elim) Used terms: bits1, boolvar1, bits