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Proof step 16 from proof lastn-fill-sgn (previous step)

n = # fillsgn(n, sgn(bits)), # sgn(bits) < # bits, # sgn(bits) < n, n < # bits, # sgn(bits) + 1 < # bits ⊦ fillsgn(n, sgn(bits)) = lastn(n, bits), bits = [], n = # bits

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 17:
n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n
Proof step 18:
∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n, n = # fillsgn(n, sgn(bits)), # sgn(bits) < # bits, # sgn(bits) < n, n < # bits, # sgn(bits) + 1 < # bits ⊦ fillsgn(n, sgn(bits)) = lastn(n, bits), bits = [], n = # bits

Rule argument

seq: n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n subst: [n, bits] → [n, bits] rotate: F spec: list inst: bitlista name: split-rest

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n → ¬ n < m; Used terms: n, bits