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Proof step 29 from proof and-size-one (previous step)

bits = bits4 + boolvar0 ', bits.first, # bits ≤ # bits3 + 1, ∀ bits2, bits1, bits0. # bits2 < # bits → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits = true ', bits = [], bits3 + boolvar ' = (bits3 + boolvar ').first ', bitsand(bits4, bits0 + bits3) = bits0 + bitsand(bits4, bits3), bits4.first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
(bits4 + boolvar0 ').first, # bits4 ≤ # bits3, ∀ bits2, bits1, bits0. # bits2 < #(bits4 + boolvar0 ') → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits4 + boolvar0 ' = true ', bits3 + boolvar ' = (bits3 + boolvar ').first ', bitsand(bits4, bits0 + bits3) = bits0 + bitsand(bits4, bits3), bits4.first

Rule argument

Additional information

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