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Proof step 12 from proof int2int-lastn-posneg (previous step)

i = bits2uint(lastn(m, int2bits(i))) + 2 ^ n→i(m) * bits2uint(bits), int2bits(i) = bits + lastn(m, int2bits(i)), # bits = # int2bits(i) - m, 0 < i, lastn(m, int2bits(i)).first, m ≤ # int2bits(i) ⊦ m = 0, i % 2 ^ n→i(m) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), bits = [], bits.first

Heuristic var elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
bits = boolvar + bits0, i = bits2uint(lastn(m, int2bits(i))) + 2 ^ n→i(m) * bits2uint(bits), int2bits(i) = bits + lastn(m, int2bits(i)), # bits = # int2bits(i) - m, 0 < i, lastn(m, int2bits(i)).first, m ≤ # int2bits(i) ⊦ m = 0, i % 2 ^ n→i(m) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), bits = [], boolvar

Rule argument

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