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Proof step 27 from proof bitsor-first (previous step)

bitsor(true ' + bits3, bits4) = false ' + bits1.butlast, boolvar ∨ boolvar0 ↔ bits1.last, (boolvar ∨ boolvar0) ∈ bits1, ∀ bits1, bits, bits2. # bits2 < #(bits4 + boolvar0 ') → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ bits1 = [], bits4 = [], (boolvar ∨ boolvar0) ∧ ¬ true ∈ bits1

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits1, bits, bits2] → [bits1.butlast, bits3, bits4]

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: bits1.butlast, bits3, bits4