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Proof step 61 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < #(false + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n0 + # bits2, false ' + bitsand(mklist(true, # bits2), bits2)) = mklist(false, n0) + bits2, n0 = 0, n0 = 1, bits2 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 62:
⊦ ¬ false ∈ bits ∧ # bits ≤ # bits0 ∧ bits ≠ [] → bitsand(bits, bits0) = bits0
Proof step 63:
¬ false ∈ mklist(true, # bits2) ∧ # mklist(true, # bits2) ≤ # bits2 ∧ mklist(true, # bits2) ≠ [], ∀ n, bits, bits1. # bits1 < #(false + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n0 + # bits2, false ' + bits2) = mklist(false, n0) + bits2, n0 = 0, n0 = 1, bits2 = []

Rule argument

name: add-ones-lelen rotate: F subst: [bits0, bits] → [bits2, mklist(true, # bits2)]

Additional information

Used simplifier rules: ⊦ ¬ false ∈ mklist(true, n);% ⊦ # mklist(boolvar, n) = n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ ¬ n < n;% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ # bits = 0 ↔ bits = []; Used terms: bits2, mklist(true, # bits2)