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Proof step 7 from proof links-add-entry (previous step)

p_ino ∈ dirs ⊦ ∀ alink. alink ∈ links(ino0, dirs[p_ino, mkdir(m0, md, n, n0, entries[str, ino0])]) ↔ alink ∈ links(ino0, dirs[p_ino, mkdir(m0, md, n, n0, entries)]) ++ mklink(p_ino, str)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
p_ino ∈ dirs ⊦ alink.pino ∈ dirs ∧ alink.name ∈ (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries[str, ino0]);dirs[alink.pino]).entries ∧ ino0 = (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries[str, ino0]);dirs[alink.pino]).entries[alink.name] ↔ alink = mklink(p_ino, str) ∨ alink.pino ∈ dirs ∧ 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]

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];% ⊦ alink ∈ ls ++ alink0 ↔ alink = alink0 ∨ alink ∈ ls; (s)% ⊦ dirs[ino, ad][ino0] = (ino0 = ino ∧ ino ∈ dirs ⊃ ad;dirs[ino0]); (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)