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Proof step 24 from proof bitsor-ones-false (previous step)

∀ bits1, bits0. # bits0 < #(bits2 + true ') → true → bits0 = [] ∨ bitsor(bits0, false ' + bits1) = bits0 ∨ false ∈ bits0 ⊦ bits2 = [], bitsor(bits2, false ' + bits) = bits2, false ∈ bits2

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits1, bits0] → [bits, bits2]

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: bits, bits2