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Proof step 7 from proof And-discard (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ st1, P. # st1 < # st0[a, d] → true → st1 = ∅ ∨ (λ a, d. P) ∧_ st1 = P ⊦ (P ∧ (λ a, d. P) ∧_ st0) = P, a ∈ st0

Rule argument

Additional information

Used simplifier rules: And-ins: ⊦ ¬ a ∈ st → adP ∧_ st[a, d] = (adP(a, d) ∧ adP ∧_ st);