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Proof step 9 from proof id-idx-freelist-cons (previous step)

# lpt1 = # lpt0, lpt0[n] = mklp(m, m0, lpf, m1), ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags, ∀ n. n < # lpt0 ∧ lpt1[n].flags = LP_FREE ↔ n ∈ freelist ⊦ n < # lpt0 ∧ lpf = LP_FREE ↔ n ∈ freelist, dups(freelist)

User applied rule all left

And obtained the following new goals

Proof step 10:
# lpt1 = # lpt0, n < # lpt0 ∧ lpt1[n].flags = LP_FREE ↔ n ∈ freelist, lpt0[n] = mklp(m, m0, lpf, m1), ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ n < # lpt0 ∧ lpf = LP_FREE ↔ n ∈ freelist, dups(freelist)

Rule argument

pos: left 4 terms: n abort: F discard: T computed: T

Additional information

Used terms: n