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Proof step 21 from proof bitsxor-zeros (previous step)

# bits3 ≤ # bits2, ∀ bits, bits1. # bits1 < #(bits3 + false ') → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ bits3 = [], bits2 = [], bitsxor(bits2, bits3) = bits2, true ∈ bits3

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits, bits1] → [bits2, bits3]

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, bits3