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Proof step 13 from proof And-forall (previous step)

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

User applied rule induction

And obtained the following new goals

Proof step 12:
(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)

Rule argument

[INDHYPARG : [STANDARDINDUCTIONTYPE : ], true, false, # st, [] → [], <]

Additional information

[[CONCRETESIMPRULES : ([CSIMPLESSPRED : <])], [EXTRATERMS : (# st)]]