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Proof step 45 from proof abs-acyclic (previous step)

tree[p] = dnode(md0, sdt), entries[str] = ROOT_INO, ad = mkdir(m, md, m0, m1, entries0), m = # entries0, md = sdt[str].meta, (( ( ((ROOT_INO => ad * abs(sdt[str].entries, entries0)) * ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries)) * abs(sdt -- str, entries -- str)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), sdt[str].dir?, abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ sdt, str ∈ entries ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
entries[str] = ROOT_INO, tree[p] = dnode(md0, sdt), (( ( ( (ROOT_INO => mkdir(# entries0, sdt[str].meta, m0, m1, entries0) * abs(sdt[str].entries, entries0)) * ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries)) * abs(sdt -- str, entries -- str)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), sdt[str].dir?, abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ entries, str ∈ sdt ⊦

Proof step 45 from proof abs-acyclic (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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