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Proof step 13 from proof bitsand-zeros-01 (previous step)

# bits2 + 1 < # bits3, ∀ bits1, bits0. # bits0 < #(bits3 + false ') → 1 + # bits1 < # bits0 → bits0 = [] ∨ bitsand(bits0, false ' + bits1) = false ' + mklist(false, # bits1) ∨ true ∈ bits0 ⊦ bits3 = [], bitsand(bits3, false ' + bits2) + false ' = false ' + false ' + mklist(false, # bits2), true ∈ bits3

User applied rule apply induction

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ #(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, bits3