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Proof step 2 from proof path-split (previous step)

∀ q, n, m, dirs, files. ispath(p0 + q, n, m, dirs, files) ↔ (∃ m0. ispath(p0, n, m0, dirs, files) ∧ ispath(q, m0, m, dirs, files)) ⊦ ispath((name + p0) + q, n, m, dirs, files) ↔ (∃ m0. ispath(name + p0, n, m0, dirs, files) ∧ ispath(q, m0, m, dirs, files))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ∈ dirs, name ∈ dirs[n].entries, ∀ q, n, m, dirs, files. ispath(p0 + q, n, m, dirs, files) ↔ (∃ m0. ispath(p0, n, m0, dirs, files) ∧ ispath(q, m0, m, dirs, files)) ⊦ ispath(p0 + q, dirs[n].entries[name], m, dirs, files) ↔ (∃ m0. ispath(p0, dirs[n].entries[name], m0, dirs, files) ∧ ispath(q, m0, m, dirs, files))

Rule argument

Additional information

Used simplifier rules: path-rec: ⊦ ispath(str ' + p, n, m, dirs, files) ↔ valid-dir-ino(n, dirs, files) ∧ str ∈ dirs[n].entries ∧ ispath(p, dirs[n].entries[str], m, dirs, files);%valid-dir-ino: ⊦ valid-dir-ino(ino, dirs, files) ↔ ino ∈ dirs;%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str + p = str ' + p; (s)