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Proof step 109 from proof inv-rmdir (previous step)

(entries0[str] = m ⊃ mkdir(m3, md3, # links(entries0[str], dirs0), m2 - 1, entries0 -- str) ;mkdir(m1, md2, # links(entries0[str], dirs0), m4, ∅) ) = mkdir(m0, md, m5, m6, entries), dirs0[entries0[str]] = mkdir(m1, md2, # links(entries0[str], dirs0), m4, ∅), dirs0[m] = mkdir(m3 + 1, md3, # links(m, dirs0), m2, entries0), m5 = 0, afs-cons(dirs0, files), entries0[str] ∈ dirs0, str ∈ entries0, m ∈ dirs0, 1 ≤ m2 ⊦ # links(m, dirs0) = 0, entries0[str] = ROOT_INO, links(m, dirs0) = ∅, m = ROOT_INO, m2 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 110:
(entries0[str] = m ⊃ mkdir(m3, md3, # links(entries0[str], dirs0), m2 - 1, entries0 -- str) ;mkdir(m1, md2, # links(entries0[str], dirs0), m4, ∅) ) = mkdir(m0, md, 0, m6, entries), dirs0[m] = mkdir(m3 + 1, md3, # links(m, dirs0), m2, entries0), dirs0[entries0[str]] = mkdir(m1, md2, # links(entries0[str], dirs0), m4, ∅), afs-cons(dirs0, files), m ∈ dirs0, str ∈ entries0, entries0[str] ∈ dirs0, 1 ≤ m2 ⊦ # links(m, dirs0) = 0, entries0[str] = ROOT_INO, links(m, dirs0) = ∅, m = ROOT_INO, m2 = 0

Proof step 109 from proof inv-rmdir (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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