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Proof step 48 from proof path-split-wand (previous step)

((ino0 => mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]) * abs(sdt0, entries0)))(dirs1), ((abs(tree, ino) * abs(sdt0, entries0) * ino0 => mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino])))(dirs2 ∪ dirs ∪ dirs1), ((abs(tree[p0], ino1) -* abs(tree, ino)))(dirs), abs(tree[p0], ino1)(dirs0), abs(tree[p0], ino1)(dirs2), abs(tree, ino)(dirs2 ∪ dirs), p0 ∈ tree, dirs0 ⊥ dirs ∪ dirs1, dirs1 ⊥ dirs, dirs2 ⊥ dirs, dirs1 ⊥ dirs2 ⊦ str ∈ sdt0, str ∈ entries0, ∃ ad. str ∈ ad.entries ∧ (((abs(tree, ad.entries[str]) * abs(sdt0, ad.entries -- str)) * ino0 => ad))(dirs0 ∪ dirs ∪ dirs1) ∧ ad.meta = md ∧ ad.size = # ad.entries

User applied rule exists right

And obtained the following new goals

Proof step 49:
((abs(tree, ino) * abs(sdt0, entries0) * ino0 => mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino])))(dirs2 ∪ dirs ∪ dirs1), ((ino0 => mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]) * abs(sdt0, entries0)))(dirs1), ((abs(tree[p0], ino1) -* abs(tree, ino)))(dirs), abs(tree, ino)(dirs2 ∪ dirs), abs(tree[p0], ino1)(dirs2), abs(tree[p0], ino1)(dirs0), p0 ∈ tree, dirs1 ⊥ dirs2, dirs2 ⊥ dirs, dirs1 ⊥ dirs, dirs0 ⊥ dirs ∪ dirs1 ⊦ str ∈ entries0, str ∈ sdt0, str ∈ mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]).entries ∧ (( ( abs(tree, mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]).entries[str]) * abs(sdt0, mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]).entries -- str)) * ino0 => mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino])) )(dirs0 ∪ dirs ∪ dirs1) ∧ mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]).meta = md ∧ mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]).size = # mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]).entries

Rule argument

pos: right 3 terms: mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]) abort: F discard: T computed: T

Additional information

Used terms: mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino])