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Proof step 8 from proof path-split-one (previous step)

dirs[m] = mkdir(n0, md, n, n1, entries), ispath(p, n, m, dirs, files), str ∈ entries, m ∈ dirs, entries[str] ∈ files ∨ entries[str] ∈ dirs ⊦ ∃ m0. ispath(p, n, m0, dirs, files) ∧ m0 ∈ dirs ∧ str ∈ dirs[m0].entries ∧ (entries[str] ∈ dirs ∨ entries[str] ∈ files) ∧ dirs[m0].entries[str] = entries[str]

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: m abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s) Used terms: m