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Proof step 2 from proof tr-null (previous step)

⊦ tr(null, t)(H) ↔ ((emp ∧ ⌜t = leaf⌝ ))(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
tr(null, t)(H) ⊦ H = ∅ ∧ t = leaf

Rule argument

Additional information

Used simplifier rules: tr-base: ⊦ tr(r, leaf) = (emp ∧ ⌜r = null⌝ );% ⊦ (true ∧ P) = P; (s)% ⊦ ⌜true⌝ = true; (s)%and-com: ⊦ (P0 ∧ P1) = (P1 ∧ P0); from specification heap-sep%and-ass: ⊦ ((P0 ∧ P1) ∧ P2) = (P0 ∧ P1 ∧ P2); from specification heap-sep% ⊦ ⌜bv⌝ (H) ↔ bv; (s)% ⊦ emp(H) ↔ H = ∅; (s)% ⊦ ((P0 ∧ P1))(H) ↔ P0(H) ∧ P1(H); (s)