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Proof step 28 from proof bitsand-app-ones (previous step)

∀ bits0, bits, bits2. # bits2 < #(bits4 + boolvar0 ') → true → bitsand(bits + true ' + bits0, false ' + bits2) = bitsand(true ' + bits0, false ' + bits2) ∨ false ∈ bits ⊦ bitsand(bits + true ' + bits3, false ' + bits4) = bitsand(true ' + bits3, false ' + bits4), false ∈ bits

User applied rule apply induction

And closed the goal!

Rule argument

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

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: bits3, bits, bits4