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Proof step 11 from proof And-forall (previous step)

(adP0 ∧_ st)(H), ∀ a, H. a ∈ st ∧ adP0(a, st[a])(H) → adP(a, st[a])(H), ∀ st0, adP0, H, adP. # st0 < # st → ((adP0 ∧_ st0)(H) ∧ (∀ a, H. a ∈ st0 ∧ adP0(a, st0[a])(H) → adP(a, st0[a])(H)) → (adP ∧_ st0)(H)) ⊦ (adP ∧_ st)(H)

User applied rule cut formula

And closed the goal!

Rule argument

[fma-arg: st = ∅]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : And-emp, , , ⊦ adP ∧_ ∅ = true], [CSIMPSEQ : ⊦ true(H)])], [EXTRAFMAS : (st = ∅)]]