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Proof step 6 from proof entries-any (previous step)

((abs(sdt, sdi) * D))(dirs) ⊦ ∀ str, tree, sdt0. ¬(sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0), ∃ tree, ino, str, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ sdi = sdi0[str, ino] ∧ ((abs(tree, ino) * abs(sdt0, sdi0) * D))(dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ ¬ name ∈ sdt → abs(sdt[name, tree], entries) = (⌜name ∈ entries⌝ ∧ abs(tree, entries[name]) * abs(sdt, entries -- name))
Proof step 8:
(((abs(tree, sdi[str]) * abs(sdt0, sdi -- str)) * D))(dirs), str ∈ sdi ⊦ str ∈ sdt0, ∃ tree0, ino, str0, sdi0, sdt1. ¬ str0 ∈ sdt1 ∧ sdt0[str, tree] = sdt1[str0, tree0] ∧ ¬ str0 ∈ sdi0 ∧ sdi = sdi0[str0, ino] ∧ ((abs(tree0, ino) * abs(sdt1, sdi0) * D))(dirs)

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));% ⊦ ⌜bv⌝ (dirs) ↔ bv; (s)% ⊦ ((D ∧ D0))(dirs) ↔ D(dirs) ∧ D0(dirs); (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]%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]% ⊦ (⌜bv⌝ ∧ D) * D0 = (⌜bv⌝ ∧ D * D0); (s)