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Proof step 35 from proof bits2int-lasteq (previous step)

bits0.first ⊦ bits0 = [], bits2uint(bitsinvert(bits1)) = bits2uint(bitsinvert(bits0.rest)) ↔ ~1 = bits2int(bits0) + bits2uint(bitsinvert(bits1))

Heuristic var elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 36:
bits0 = boolvar + bits, boolvar ⊦ bits0 = [], bits2uint(bitsinvert(bits1)) = bits2uint(bitsinvert(bits)) ↔ ~1 = bits2int(bits0) + bits2uint(bitsinvert(bits1))

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