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Proof step 3 from proof tr-null (previous step)

tr(null, t)(H) ⊦ H = ∅ ∧ t = leaf

User applied rule constructor cut

And obtained the following new goals

Proof step 4:
t = branch(t0, a, t1), tr(null, t)(H) ⊦ H = ∅ ∧ t = leaf

Rule argument

var: t

Additional information

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