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Proof step 19 from proof and-size-one (previous step)

bits1 = bits1.first ', # bits3 < # bits1, ∀ bits2, bits1, bits0. # bits2 < #(true + bits3) → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bitsand(true ' + bits3.butlast, bits0 + bits1.butlast) + (bits1.first ∧ bits3.last)' = bits0 + bitsand(true ' + bits3, bits1), bits1 = [], # bits0 + # bits1 = 0, bits3 = [], bits0 = []

Heuristic elimination 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] → [bits1, boolvar, bits] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: ⊦ boolvar + bits = boolvar ' + bits;% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ m + 1 ≠ 0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ bits + [] = bits;% ⊦ boolvar '.butlast = [];% ⊦ # [] = 0;% ⊦ m ≤ 0 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = [];%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits1, boolvar, bits