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Proof step 27 from proof and-zerosin (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar '.first ↔ boolvar;% ⊦ [] + bits = bits;% ⊦ #(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