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Proof step 10 from proof And-discard (previous step)

(λ a, d. P) ∧_ st0 = P, ∀ st1, P. # st1 < # st0[a, d] → true → st1 = ∅ ∨ (λ a, d. P) ∧_ st1 = P ⊦ (P ∧ P) = P, st0 = ∅, a ∈ st0

User applied rule expand right

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ((P0 ∧ P1))(H) ↔ P0(H) ∧ P1(H); (s)%and-ass: ⊦ ((P0 ∧ P1) ∧ P2) = (P0 ∧ P1 ∧ P2); from specification heap-sep%and-com: ⊦ (P0 ∧ P1) = (P1 ∧ P0); from specification heap-sep