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Proof step 5 from proof px-parent (previous step)

tree[p].dir? ↔ m ∈ dirs, abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), ino ∈ dirs ⊦ px-parent(user, p, tree, abs-files(files)) ↔ px-parent(user, p, ino, dirs, files)

User applied rule cut formula

And obtained the following new goals

Proof step 6:
tree[p].dir? ↔ m ∈ dirs, abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), ino ∈ dirs ⊦ px-parent(user, p, tree, abs-files(files)) ↔ px-parent(user, p, ino, dirs, files), p = ε

Rule argument

fma: p = ε

Additional information

Used simplifier rules: ⊦ tree[ε] = tree; (s)% ⊦ px-parent(user, ε, tree, fs); (s)% ⊦ px-parent(user, ε, ino, dirs, files); (s) Used formulas: p = ε