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Proof step 724 from proof bitsand-cast (previous step)

# bits1 = # bitsand(bits1, false ' + true ' + bits), 2 + # bits = # bits1, sgn(mklist(false, m) + false ' + true ' + bits) = false ' + true ' + bits, m < # bits3 ⊦ mklist(false, m) + bitsand(bits1, false ' + true ' + bits) = lastn(2 + # bits + m, fillsgn(2 + # bits + m, bitsand(bits2 + mklist(false, m), bits3) + bitsand(false ' + true ' + bits, bits1))), bits1 = [], bits2 + mklist(false, m) = [], bits3 = [], bits3 + bits1 = []

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 725:
n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n
Proof step 726:
∃ bits0, bits1. bits3 = bits0 + bits1 ∧ # bits1 = m, # bits1 = # bitsand(bits1, false ' + true ' + bits), 2 + # bits = # bits1, sgn(mklist(false, m) + false ' + true ' + bits) = false ' + true ' + bits, m < # bits3 ⊦ mklist(false, m) + bitsand(bits1, false ' + true ' + bits) = lastn(2 + # bits + m, fillsgn(2 + # bits + m, bitsand(bits2 + mklist(false, m), bits3) + bitsand(false ' + true ' + bits, bits1))), bits1 = [], bits2 + mklist(false, m) = [], bits3 = [], bits3 + bits1 = []

Rule argument

seq: n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n subst: [n, bits] → [m, bits3] rotate: F spec: list inst: bitlista name: split-rest

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n → ¬ n < m; Used terms: m, bits3