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Proof step 37 from proof bitsand-fillsgn (previous step)

1 + # bits2 < n, # bits2 < n, ∀ n, bits, bits1. # bits1 < #(false + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ mklist(false, n - # bits2) + mklist(false, # bits2) = mklist(false, n)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, # bits2, n0] rotate: T spec: nat inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # bits2, n0