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Proof step 95 from proof and-sgn (previous step)

(¬ boolvar0) ∈ bits0, (¬ boolvar0 ∨ ¬ boolvar) ∈ bitsand(bits0, bits2), (¬ boolvar0 ∨ ¬ boolvar) ∈ bitsand(sgn(bits0), bits2), ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits2 = [], bits0 + boolvar0 ' = bits0.first ', bits2 + boolvar ' = bits2.first ', bits0 = [], sgn(bitsand(sgn(bits0), bits2)) = sgn(bitsand(bits0, bits2))

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits1, bits] → [bits2, bits0]

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, bits0