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Proof step 3 from proof path-uni (previous step)

ispath(p, n0, n1, dirs0, files), dirs0 ⊥ dirs1 ⊦ ispath(p, n0, n1, dirs0 ∪ dirs1, files)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
ispath(name + p0, n0, n1, dirs0, files), dirs0 ⊥ dirs1, ∀ n0, n1, dirs0, files, dirs1. ispath(p0, n0, n1, dirs0, files) ∧ dirs0 ⊥ dirs1 → ispath(p0, n0, n1, dirs0 ∪ dirs1, files) ⊦ ispath(name + p0, n0, n1, dirs0 ∪ dirs1, 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, +;;% ⊦ dirs ⊥ dirs0 → (ino ∈ dirs ∪ dirs0 ↔ ino ∈ dirs ∨ ino ∈ dirs0); (s)