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Proof step 40 from proof bitsor-bi-zeroes (previous step)

bits3 = boolvar + bits, # bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2uint((bits3 + bitsor(bits0, bits2)) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits2 = [], bits0 = [], bits3 = [], (bits3 + bitsor(bits0, bits2)) + bits1 = [], true ∈ bits3, true ∈ bits0, boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
# bits0 = # bits2 ⊦ bits2 = [], bits0 = [], bits2uint(bits + bitsor(bits0, bits2) + bits1) = bits2uint(bits2 + bits1), bits = [], true ∈ bits0, true ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ boolvar ∈ boolvar0 ' + bits ↔ ¬(boolvar ↔ boolvar0) ∧ ¬ boolvar ∈ bits;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits2uint(false ' + bits) = bits2uint(bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;