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Proof step 39 from proof bitsor-bi-zeroes (previous step)

# bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2uint((bits3 + bitsor(bits0, bits2)) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits2 = [], bits0 = [], bits3 = [], (bits3 + bitsor(bits0, bits2)) + bits1 = [], true ∈ bits3, true ∈ bits0, bits3.first

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 40:
bits3 = boolvar + bits, # bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2uint((bits3 + bitsor(bits0, bits2)) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits2 = [], bits0 = [], bits3 = [], (bits3 + bitsor(bits0, bits2)) + bits1 = [], true ∈ bits3, true ∈ bits0, boolvar

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits3, boolvar, bits] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits3, boolvar, bits