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Proof step 37 from proof abs-nlinks-h (previous step)

tree[p0 + str '] = fnode(⌜ino1⌝ ), tree[p0] = dnode(md, sdt), ((m0 => mkdir(# entries, md, m1, m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * abs|(tree, ino0, p0, m0)) )(dirs), ispath(p0, ino0, m0, dirs, files), str ∈ entries, ino0 ∈ dirs, ino1 ∈ files, p0 ∈ tree, str ∈ sdt, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ alink. alink ∈ links(ino1, dirs) ↔ alink ∈ ∅, entries[str] ∈ dirs ∨ entries[str] ∈ files ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
tree[p0] = dnode(md, sdt), tree[p0 + str '] = fnode(⌜ino1⌝ ), ((m0 => mkdir(# entries, md, m1, m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * abs|(tree, ino0, p0, m0)) )(dirs), ispath(p0, ino0, m0, dirs, files), str ∈ sdt, p0 ∈ tree, ino1 ∈ files, ino0 ∈ dirs, str ∈ entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ alink. ¬ alink.pino ∈ dirs ∨ ¬ alink.name ∈ dirs[alink.pino].entries ∨ ino1 ≠ dirs[alink.pino].entries[alink.name], entries[str] ∈ dirs ∨ entries[str] ∈ files ⊦

Rule argument

Additional information

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