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Proof step 2 from proof entries-fold-p (previous step)

((abs(tree, ino) * abs(sdt, sdi) * D))(dirs) ⊦ ((abs(sdt[str, tree], sdi[str, ino]) * D))(dirs), str ∈ sdi, str ∈ sdt

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ ¬ name ∈ sdt → abs(sdt[name, tree], entries) = (⌜name ∈ entries⌝ ∧ abs(tree, entries[name]) * abs(sdt, entries -- name))
Proof step 4:
⊦ ((abs(sdt, sdi) * D))(dirs) ∧ ¬ str ∈ sdt → sdi -- str = sdi

Rule argument

Additional information

Used simplifier rules: abs-rec: ⊦ ¬ name ∈ sdt → abs(sdt[name, tree], entries) = ( ⌜name ∈ entries⌝ ∧ abs(tree, entries[name]) * abs(sdt, entries -- name));%entries-del: ⊦ ((abs(sdt, sdi) * D))(dirs) ∧ ¬ str ∈ sdt → sdi -- str = sdi;%and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]% ⊦ (true ∧ D) = D; (s)%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]% ⊦ entries[str, ino] -- str = entries -- str; (s)% ⊦ entries[str, ino][str] = ino; (s)% ⊦ ⌜true⌝ = true; (s)% ⊦ str ∈ entries[str, ino]; (s)