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Proof step 21 from proof bits2int-range (previous step)

bits = boolvar + bits0, # bits = m + 1 ⊦ bits = [], boolvar, ¬ 2 ^ n→i(m) + bits2uint(bits) < 0 ∧ ¬ 2 ^ n→i(m) ≤ bits2uint(bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
bits2uint(bits0) + 2 ^ n→i(# bits0) < 0 ⊦

Rule argument

Additional information

Used simplifier rules: bits2uint-zeros-03: ⊦ bits2uint(false ' + bits) = bits2uint(bits);%bits2uint-nle: ⊦ ¬ 2 ^ n→i(# bits) ≤ bits2uint(bits);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ boolvar ' + bits ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ boolvar + bits = boolvar ' + bits;