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Proof step 32 from proof bitsand-first-samesize (previous step)

# bits2 = # bits3, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < #(bits3 + boolvar2 ') → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bits3 = [], bitsand(boolvar ' + bits2, boolvar0 ' + bits3) = (boolvar ∧ boolvar0)' + bitsand(bits2, bits3)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [boolvar0, boolvar, bits1, bits0] → [boolvar0, boolvar, bits2, 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: boolvar0, boolvar, bits2, bits3