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Proof step 31 from proof lastn-fill-sgn (previous step)

# bits0 = # fillsgn(# bits0, sgn(true ' + bits0)), 1 + # sgn(true ' + bits0) < # bits1 + # bits0, # sgn(true ' + bits0) < # bits0, # sgn(true ' + bits0) < # bits1 + # bits0, true ∈ bits1 ⊦ fillsgn(# bits0, sgn(true ' + bits0)) = bits0, bits0 = [], bits1 = [], bits1 + bits0 = [], false ∈ bits1, bits0.first

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] → [bits0, boolvar, bits] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: sgn-base-01: ⊦ sgn(true ' + false ' + bits) = true ' + false ' + bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ bits + boolvar ' + bits0 ≠ [];% ⊦ m + n < m + n0 ↔ n < n0;%n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits0, boolvar, bits