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Proof step 33 from proof bitsor-com (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bitsor(bits, bits1) = bitsor(bits1, bits) ⊦ bits2 = [], bits3 = [], bitsor(bits2, bits3) + (boolvar ∨ boolvar0)' = bitsor(bits3, bits2) + (boolvar0 ∨ boolvar)'

Rule argument

Additional information

Used simplifier rules: or-last: ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsor(bits + boolvar ', bits0 + boolvar0 ') = bitsor(bits, bits0) + (boolvar ∨ boolvar0)';% ⊦ 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;