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Proof step 5 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 obtained the following new goals

Proof step 6:
((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) ⊦ st = ∅, ((adP ∧_ st))(H)

Rule argument

fma: st = ∅

Additional information

Used simplifier rules: And-emp: ⊦ adP ∧_ ∅ = true;% ⊦ true(H); (s) Used formulas: st = ∅