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Proof step 17 from proof bitsand-zeros (previous step)

bits = bits3 + boolvar1 ', # bits ≤ 2 + # bits2 ∨ boolvar, ∀ boolvar, bits1, bits0. # bits0 < # bits → # bits0 ≤ 1 + # bits1 ∨ boolvar → bits0 = [] ∨ bitsand(bits0, boolvar ' + bits1) = bits0 ∨ true ∈ bits0 ⊦ # bits = 1, bits = [], bitsand(bits3, boolvar ' + bits2) + (boolvar1 ∧ boolvar0)' = bits, true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# bits3 ≤ # bits2 + 1 ∨ boolvar, ∀ boolvar, bits1, bits0. # bits0 < #(bits3 + false ') → # bits0 ≤ 1 + # bits1 ∨ boolvar → bits0 = [] ∨ bitsand(bits0, boolvar ' + bits1) = bits0 ∨ true ∈ bits0 ⊦ bits3 = [], bitsand(bits3, boolvar ' + bits2) = bits3, true ∈ bits3

Rule argument

Additional information

Used simplifier rules: ⊦ bits + bits0 = bits1 + bits0 ↔ bits = bits1;%n < n0 ⊦ m + (* n) ≤ m0 + (* n0) ↔ m ≤ m0 + (n0 - n);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ bits + boolvar ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;