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Proof step 50 from proof bitsand-fillsgn (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 51:
n = n0 + # bits2 ∧ # bits2 ≤ n, # bits2 < n, 1 + # bits2 < n, ∀ n, bits, bits1. # bits1 < #(false + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(mklist(true, n), false ' + bits2)) = mklist(false, n0) + bits2

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;%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # bits2, n0