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Proof step 3 from proof path-in-end (previous step)

ispath(p, n, m, dirs, files) ⊦ m ∈ files, m ∈ dirs

User applied rule structural induction

And obtained the following new goals

Proof step 4:
ispath(name + p0, n, m, dirs, files), ∀ n, m, dirs, files. ispath(p0, n, m, dirs, files) → m ∈ files ∨ m ∈ dirs ⊦ m ∈ dirs, m ∈ files

Rule argument

indvar: p

Additional information

Used simplifier rules: path-here: ⊦ ispath(ε, n, m, dirs, files) ↔ valid-ino(n, dirs, files) ∧ m = n;%valid-ino: ⊦ valid-ino(ino, dirs, files) ↔ ino ∈ dirs ∨ ino ∈ files;%path freely generated by ε, +;;