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Proof step 47 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < #(boolvar + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ n = 0, fillsgn(n, bitsand(mklist(true, n), boolvar ' + bits2)) = fillsgn(n, boolvar ' + bits2), ¬ n ≤ # bits2 ∧ boolvar

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 48:
⊦ ¬ false ∈ bits ∧ # bits ≤ # bits0 ∧ bits ≠ [] → bitsand(bits, bits0) = bits0
Proof step 49:
¬(¬ false ∈ mklist(true, n) ∧ # mklist(true, n) ≤ #(boolvar ' + bits2) ∧ mklist(true, n) ≠ []), ∀ n, bits, bits1. # bits1 < #(boolvar + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ n = 0, fillsgn(n, bitsand(mklist(true, n), boolvar ' + bits2)) = fillsgn(n, boolvar ' + bits2), ¬ n ≤ # bits2 ∧ boolvar

Rule argument

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

Additional information

Used terms: boolvar ' + bits2, mklist(true, n)