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Proof step 58 from proof and-ass (previous step)

∀ bits2, bits0, bits1. # bits2 < #(bits4 + boolvar0 ') → true → bits2 = [] ∨ bits0 = [] ∨ bits1 = [] ∨ bitsand(bits2, bitsand(bits0, bits1)) = bitsand(bitsand(bits2, bits0), bits1) ⊦ bitsand(bits4, bitsand(bits, bits3)) + (boolvar0 ∧ boolvar1 ∧ boolvar)' = bitsand(bitsand(bits4, bits), bits3) + ((boolvar0 ∧ boolvar1) ∧ boolvar)', bitsand(bits4, bits) = [], bits3 = [], bits4 = [], bits4 + boolvar0 ' = bits4.first ', bitsand(bits, bits3) = [], bits + boolvar1 ' = (bits + boolvar1 ').first ', bits3 + boolvar ' = bits3.first '

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bits2, bits0, bits1] → [bits4, bits, bits3]

Additional information

Used simplifier rules: ⊦ bits + boolvar ' = bits0 + boolvar0 ' ↔ bits = bits0 ∧ (boolvar ↔ boolvar0);% ⊦ [] + bits = bits;% ⊦ boolvar '.first ↔ boolvar;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ #(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: bits4, bits, bits3