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Proof step 44 from proof or-invandinv (previous step)

∀ bits1, bits0. # bits1 < #(bits2 + boolvar ') → true → bits1 = [] ∨ bits0 = [] ∨ bitsor(bits1, bits0) = bitsinvert(bitsand(bitsinvert(bits1), bitsinvert(bits0))) ⊦ bitsor(bits2, bits) = bitsinvert(bitsand(bitsinvert(bits2), bitsinvert(bits))), bits2 = [], bits2 + boolvar ' = bits2.first ', bits = [], bits + boolvar0 ' = bits.first '

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits1, bits0] → [bits2, bits]

Additional information

Used simplifier rules: ⊦ #(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, bits