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Proof step 12 from proof bitsand-zeros-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# bits2 + 1 < # bits3, ∀ bits1, bits0. # bits0 < #(bits3 + false ') → 1 + # bits1 < # bits0 → bits0 = [] ∨ bitsand(bits0, false ' + bits1) = false ' + mklist(false, # bits1) ∨ true ∈ bits0 ⊦ bits3 = [], bitsand(bits3, false ' + bits2) + false ' = false ' + false ' + mklist(false, # bits2), true ∈ bits3

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);%n0 < n ⊦ m + (* n) < m0 + (* n0) ↔ m + (n - n0) < m0;% ⊦ 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;