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Proof step 10 from proof int2int-lastn-posneg (previous step)

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

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

Additional information

Used simplifier rules: int2bits-first-03: ⊦ 0 < j → int2bits(j) ≠ true ' + bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar, bits0