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Proof step 3 from proof bitsand-len-true (previous step)

bits0 = boolvar + bits1, # bits0 ≤ # bits, boolvar ⊦ bits0 = [], # bitsand(bits, bits0) = # bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# bits1 < # bits ⊦ # bitsand(bits, true ' + bits1) = # bits, bits = []

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 + 1 ≤ n ↔ m < n;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;