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Proof step 67 from proof and-sgn (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits1, bits] → [bits2, bits0]

Additional information

Used simplifier rules: and-onel-01: ⊦ bits ≠ [] → bitsand(true ', bits) = bits;% ⊦ ¬ false ∈ bits ∧ bits ≠ [] → sgn(bits) = true ';% ⊦ #(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, bits0