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Proof step 20 from proof bitsand-zero-in (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits0, bits1] → [bits, bits2]

Additional information

Used simplifier rules: and-onesin: ⊦ ¬ false ∈ bits ∧ bits ≠ [] ∧ bits0 ≠ [] → (boolvar ∈ bitsand(bits, bits0) ↔ boolvar ∈ bits0);% ⊦ #(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