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Proof step 72 from proof bits2int-pos (previous step)

#(boolvar ' + bits0) ≤ 8 ⊦ 0 ≤ bytes2int(i2b(bits2int(boolvar ' + bits0)) ') ↔ ¬ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
# bits0 ≤ 7 ⊦ 0 ≤ b2i(i2b(bits2int(boolvar ' + bits0))) ↔ ¬ boolvar

Rule argument

Additional information

Used simplifier rules: bytes2int-one: ⊦ bytes2int(by ') = b2i(by);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ m + (* n) ≤ (* n0) ↔ m ≤ n0 - n;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;