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Proof step 2 from proof bits2int-split-01 (previous step)

bits.first ⊦ ~1 = 2 ^ n→i(# bits0) * bits2uint(bitsinvert(bits)) + bits2uint(bitsinvert(bits0)) + bits2int(bits + bits0), bits = [], bits0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
bits = boolvar + bits1, boolvar ⊦ ~1 = 2 ^ n→i(# bits0) * bits2uint(bitsinvert(bits)) + bits2uint(bitsinvert(bits0)) + bits2int(bits + bits0), bits = [], bits0 = []

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits, boolvar, bits1] 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: bits, boolvar, bits1