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Proof step 10 from proof uint2bits-2pot-05 (previous step)

uint2bits(i) = false ' + bits, uint2bits(i) = false ' + true ' + bits.rest ↔ 2 ^ (1 + n→i(# bits - 1)) = i + 1, 0 ≤ i, false ∈ uint2bits(i) ⊦ bits = [], 2 ^ n→i(# bits) = i + 1, false ∈ 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-1

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ 0 + m = m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ boolvar ' + bits ≠ [];% ⊦ n→i(m + n) = n→i(m) + n→i(n);%⊦ ¬ boolvar ∈ boolvar0 ' + bits ↔ ¬(boolvar ↔ boolvar0) ∧ ¬ boolvar ∈ bits;%⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0); (elim) Used terms: bits, boolvar, bits0