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Proof step 3 from proof entries-empty (previous step)

⊦ ∀ D1, dirs. ((abs(sdt, ∅) * D1))(dirs) ↔ (((⌜sdt = ∅⌝ ∧ emp) * D1))(dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ abs(∅, entries) = (⌜entries = ∅⌝ ∧ emp)
Proof step 5:
((abs(sdt, ∅) * D))(dirs) ⊦ sdt = ∅ ∧ D(dirs)

Rule argument

Additional information

Used simplifier rules: abs-empty: ⊦ abs(∅, entries) = (⌜entries = ∅⌝ ∧ emp);% ⊦ (true ∧ D) = D; (s)% ⊦ ⌜true⌝ = true; (s)% ⊦ ⌜bv⌝ (dirs) ↔ bv; (s)% ⊦ ((D ∧ D0))(dirs) ↔ D(dirs) ∧ D0(dirs); (s)%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]%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]% ⊦ emp * D = D; (s)% ⊦ (⌜bv⌝ ∧ D) * D0 = (⌜bv⌝ ∧ D * D0); (s)