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Proof step 6 from proof fill-sgn-01 (previous step)

∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ fillsgn(n1 + # bits0, sgn(boolvar ' + bits0)) = mklist(boolvar, n1) + bits0, n1 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ # bits ≤ n ∧ bits ≠ [] → fillsgn(n, bits) = mklist(bits.first, n - # bits) + bits
Proof step 8:
¬(# sgn(boolvar ' + bits0) ≤ n1 + # bits0 ∧ sgn(boolvar ' + bits0) ≠ []), ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ fillsgn(n1 + # bits0, sgn(boolvar ' + bits0)) = mklist(boolvar, n1) + bits0, n1 = 0
Proof step 11:
# sgn(boolvar ' + bits0) ≤ n1 + # bits0 ∧ sgn(boolvar ' + bits0) ≠ [], ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ mklist(sgn(boolvar ' + bits0).first, n1 + # bits0 - # sgn(boolvar ' + bits0)) + sgn(boolvar ' + bits0) = mklist(boolvar, n1) + bits0, n1 = 0

Rule argument

name: fill-fill rotate: F subst: [bits, n] → [sgn(boolvar ' + bits0), n1 + # bits0]

Additional information

Used terms: sgn(boolvar ' + bits0), n1 + # bits0