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Proof step 15 from proof bits2bytes-fill (previous step)

m * 8 = 0, # bits < 8 ⊦ bits = [], i2b(bits2int(bits)) = i2b(bits2int(mklist(bits.first, 8 - # bits) + bits)), bits2int(bits) = bits2int(mklist(bits.first, 8 - # bits) + bits)

User 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: ⊦ boolvar + bits = boolvar ' + bits;% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);%⊦ bits2int(boolvar ' + mklist(boolvar, n) + bits) = bits2int(boolvar ' + bits);%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar, bits0