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Proof step 31 from proof bitsor-first-02 (previous step)

bitsor(false ' + bits3, false ' + bits0) = true ' + bits1.butlast, boolvar ∨ boolvar0 ↔ bits1.last, ∀ bits1, bits2, bits. # bits < #(bits3 + boolvar ') → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ bits1 = [], ¬ boolvar ∧ ¬ boolvar0 ∧ ¬ false ∈ bits1

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits1, bits2, bits] → [bits1.butlast, bits0, 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: bits1.butlast, bits0, bits3