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Proof step 6 from proof bits2int-pos (previous step)

¬ #(boolvar ' + bits0) ≤ 8 ⊦ 0 ≤ bytes2int(bits2bytes(boolvar ' + bits0)) ↔ ¬ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
7 < # bits0 ⊦ 0 ≤ bytes2int(bits2bytes(boolvar ' + bits0)) ↔ ¬ boolvar, bits0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ m ≤ n ↔ n < m;%n < n0 ⊦ m + (* n) ≤ (* n0) ↔ m ≤ n0 - n;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;