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Proof step 35 from proof fill-sgn-01 (previous step)

n1 + # bits0 = 1 + n, ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ boolvar ' + mklist(boolvar, n) = mklist(boolvar, n1) + bits0, n1 = 0, (¬ boolvar) ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 36:
⊦ ¬ boolvar ∈ bits ↔ mklist(¬ boolvar, # bits) = bits
Proof step 37:
n1 + # bits0 = 1 + n, ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ boolvar ' + mklist(boolvar, n) = mklist(boolvar, n1) + bits0, n1 = 0, mklist(¬ ¬ boolvar, # bits0) ≠ bits0

Rule argument

seq: ⊦ ¬ boolvar ∈ bits ↔ mklist(¬ boolvar, # bits) = bits subst: [bits, boolvar] → [bits0, ¬ boolvar] rotate: F spec: bitlista inst: name: mklist-intro

Additional information

Used terms: bits0, ¬ boolvar