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Proof step 42 from proof bitsor-mklist-in (previous step)

∀ bits, boolvar0, boolvar, n, bits1. # bits1 < #(bits3 + boolvar2 ') → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ bits3 = [], n = 0, boolvar ∈ bitsor(mklist(boolvar0, n) + bits2, bits3) ↔ boolvar ∈ bitsor(boolvar0 ' + bits2, bits3), boolvar ↔ boolvar1 ∨ boolvar2

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits, boolvar0, boolvar, n, bits1] → [bits2, boolvar0, boolvar, n, bits3]

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: bits2, boolvar0, boolvar, n, bits3