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Proof step 11 from proof id-idx-freelist-cons (previous step)

# lpt1 = # lpt0, n < # lpt0 ∧ lpf0 = LP_FREE ↔ n ∈ freelist, lpt0[n] = mklp(m, m0, lpf, m1), lpt1[n] = mklp(m2, m3, lpf0, m4), ∀ 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)

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ mklp(n2, n3, lpf, m).ref-size = n2; (s)% ⊦ mklp(n2, n3, lpf, m).size = n3; (s)% ⊦ mklp(n2, n3, lpf, m).flags = lpf; (s) Used terms: n