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Proof step 21 from proof bitsand-app-true (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

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