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Proof step 29 from proof links-add-entry (previous step)

p_ino ∈ dirs, ¬ str ∈ entries ∨ entries[str] ≠ ino0 ⊦ ino = ino0, ∀ alink. alink ∈ links(ino0, dirs[p_ino, mkdir(m0, md, n, n0, entries[str, ino])]) ↔ alink ∈ links(ino0, dirs[p_ino, mkdir(m0, md, n, n0, entries)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
p_ino ∈ dirs, alink.pino ∈ dirs, ¬ str ∈ entries ∨ entries[str] ≠ ino0 ⊦ alink.name ∈ (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries[str, ino]);dirs[alink.pino]).entries ∧ ino0 = (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries[str, ino]);dirs[alink.pino]).entries[alink.name] ↔ alink.name ∈ (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries);dirs[alink.pino]).entries ∧ ino0 = (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries);dirs[alink.pino]).entries[alink.name], ino = ino0

Rule argument

Additional information

Used simplifier rules: links: ⊦ alink ∈ links(ino, dirs) ↔ alink.pino ∈ dirs ∧ alink.name ∈ dirs[alink.pino].entries ∧ ino = dirs[alink.pino].entries[alink.name];% ⊦ dirs[ino, ad][ino0] = (ino0 = ino ∧ ino ∈ dirs ⊃ ad;dirs[ino0]); (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)