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Proof step 14 from proof And-forall (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
(adP0 ∧_ st)(H), ∀ a, H. a ∈ st ∧ adP0(a, st[a])(H) → adP(a, st[a])(H) ⊦ (adP ∧_ st)(H)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]