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Proof step 31 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])))(dirs0 ∪ dirs1), ((abs(tree[p0], ino1) * (abs(tree[p0], ino1) -* abs(tree, ino))))(dirs0), abs(tree, ino)(dirs0), p0 ∈ tree, dirs0 ⊥ dirs1 ⊦ str ∈ sdt0, str ∈ entries0, ∃ ino. (( abs(tree[p0], ino) * ( abs(tree[p0], ino) -* _∃ (λ ad. (⌜str ∈ ad.entries⌝ ∧ (abs(tree, ad.entries[str]) * abs(sdt0, ad.entries -- str)) * ino0 => ad) ∧ ⌜ad.meta = md⌝ ∧ ⌜ad.size = # ad.entries⌝ ))) )(dirs0 ∪ dirs1)

User applied rule exists right

And obtained the following new goals

Proof step 32:
((abs(tree[p0], ino1) * (abs(tree[p0], ino1) -* abs(tree, ino))))(dirs0), ((abs(tree, ino) * abs(sdt0, entries0) * ino0 => mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino])))(dirs0 ∪ dirs1), ((ino0 => mkdir(# entries0 + 1, md, m0, m1, entries0[str, ino]) * abs(sdt0, entries0)))(dirs1), abs(tree, ino)(dirs0), p0 ∈ tree, dirs0 ⊥ dirs1 ⊦ (( 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⌝ ))) )(dirs0 ∪ dirs1), str ∈ entries0, str ∈ sdt0

Rule argument

pos: right 3 terms: ino1 abort: F discard: T computed: F

Additional information

Used terms: ino1