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Proof step 17 from proof and-size-one-01 (previous step)

# bits3 = # bits2, ∀ bits, bits1, bits0. # bits < #(bits3 + boolvar ') → # bits = # bits1 → bits1 = [] ∨ bits0 = [] ∨ bitsand(true ' + bits, bits0 + bits1) = bits0 + bitsand(bits, bits1) ⊦ bits0 = [], bitsand(true ' + bits3, bits0 + bits2) = bits0 + bitsand(bits3, bits2), bits3 = [], bits2 = []

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits, bits1, bits0] → [bits3, bits2, bits0]

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: bits3, bits2, bits0