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Proof step 18 from proof bitsand-zeros (previous step)

# bits3 ≤ # bits2 + 1 ∨ boolvar, ∀ boolvar, bits1, bits0. # bits0 < #(bits3 + false ') → # bits0 ≤ 1 + # bits1 ∨ boolvar → bits0 = [] ∨ bitsand(bits0, boolvar ' + bits1) = bits0 ∨ true ∈ bits0 ⊦ bits3 = [], bitsand(bits3, boolvar ' + bits2) = bits3, true ∈ bits3

User applied rule apply induction

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ m < n → (n ≤ m ↔ false);% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ #(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: boolvar, bits2, bits3