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Proof step 35 from proof path-split-wand (previous step)

((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), p0 ∈ tree, dirs2 ⊥ dirs, dirs1 ⊥ dirs2, dirs1 ⊥ dirs ⊦ (( abs(tree[p0], ino1) * ( abs(tree[p0], ino1) -* _∃ (λ ad. (⌜str ∈ ad.entries⌝ ∧ (abs(tree, ad.entries[str]) * abs(sdt0, ad.entries -- str)) * ino0 => ad) ∧ ⌜ad.meta = md⌝ ∧ ⌜ad.size = # ad.entries⌝ ))) )((dirs2 ∪ dirs) ∪ dirs1), str ∈ entries0, str ∈ sdt0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: heap, inst: dir-heap name: uni-ass seq: ⊦ dirs ⊥ dirs0 ∧ dirs0 ⊥ dirs1 ∧ dirs1 ⊥ dirs → (dirs ∪ dirs0) ∪ dirs1 = dirs ∪ dirs0 ∪ dirs1 subst: [dirs1, dirs0, dirs] → [dirs1, dirs, dirs2] flags: <all paths>

Additional information

Used simplifier rules: ⊦ dirs ⊥ dirs0 → dirs0 ⊥ dirs; (s) Used terms: dirs1, dirs, dirs2