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Proof step 34 from proof bitsor-com (previous step)

∀ 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)'

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits, bits1] → [bits2, bits3]

Additional information

Used simplifier rules: ⊦ bits + boolvar ' = bits0 + boolvar0 ' ↔ bits = bits0 ∧ (boolvar ↔ boolvar0);% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits2, bits3