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Proof step 8 from proof bits2int-zero-02 (previous step)

bits = boolvar + bits0, bits2uint(bits) = 2 ^ n→i(# bits), boolvar, true ∈ bits ⊦ bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
bits2uint(true ' + bits0) = 2 ^ (1 + n→i(# bits0)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ boolvar + bits = boolvar ' + bits;