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Proof step 12 from proof bitsand-len-true (previous step)

bits1 = bits3 + boolvar ', # bits1 ≤ # bits2, ∀ bits, bits0. # bits0 < # bits1 → # bits0 < # bits → # bitsand(bits, true ' + bits0) = # bits ∨ bits = [] ⊦ bits2 = [], bits1 = [], # bitsand(bits2, true ' + bits3) = # bits2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# bits3 < # bits2, ∀ bits, bits0. # bits0 < #(bits3 + boolvar ') → # bits0 < # bits → # bitsand(bits, true ' + bits0) = # bits ∨ bits = [] ⊦ # bitsand(bits2, true ' + bits3) = # bits2, bits2 = []

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;% ⊦ bits + boolvar ' ≠ [];