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Proof step 57 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)(dirs0), abs(tree[p0], ino1)(dirs2), abs(tree, ino)(dirs2 ∪ dirs), p0 ∈ tree, dirs0 ⊥ dirs, dirs0 ⊥ dirs1, dirs1 ⊥ dirs, dirs2 ⊥ dirs, dirs1 ⊥ dirs2, dirs ⊥ dirs1, dirs0 ⊥ dirs → abs(tree, ino)(dirs0 ∪ dirs) ⊦ (((abs(tree, ino) * abs(sdt0, entries0)) * ino0 => mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino])))(dirs0 ∪ dirs ∪ dirs1), str ∈ sdt0, str ∈ entries0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
((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, ino)(dirs2 ∪ dirs), abs(tree[p0], ino1)(dirs2), abs(tree[p0], ino1)(dirs0), abs(tree, ino)(dirs0 ∪ dirs), p0 ∈ tree, dirs1 ⊥ dirs2, dirs2 ⊥ dirs, dirs1 ⊥ dirs, dirs0 ⊥ dirs1, dirs0 ⊥ dirs ⊦ (((abs(tree, ino) * abs(sdt0, entries0)) * ino0 => mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino])))(dirs0 ∪ dirs ∪ dirs1), str ∈ entries0, str ∈ sdt0

Rule argument

Additional information

Used simplifier rules: ⊦ dirs ⊥ dirs0 → dirs0 ⊥ dirs; (s)