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Proof step 28 from proof and-onesin (previous step)

∀ bits1, bits0, boolvar. # bits1 < #(bits3 + true ') → true → bits1 = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(bits1, bits0) ↔ boolvar ∈ bits0) ∨ false ∈ bits1 ⊦ bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', bits3 + true ' = (bits3 + true ').first ', boolvar ∈ bitsand(bits3, bits2) ∨ (boolvar ↔ boolvar0) ↔ (boolvar ↔ boolvar0) ∨ boolvar ∈ bits2, false ∈ bits3

User applied rule apply induction

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ ¬ boolvar ∈ [];% ⊦ 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, boolvar