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Proof step 9 from proof And-forall (previous step)

adP0(a, d)(H), ((adP0 ∧_ st0))(H), ∀ a0, H. a0 ∈ st0[a, d] ∧ adP0(a0, st0[a, d][a0])(H) → adP(a0, st0[a, d][a0])(H), ∀ st1, adP0, H, adP. # st1 < # st0[a, d] → ((adP0 ∧_ st1))(H) ∧ (∀ a, H. a ∈ st1 ∧ adP0(a, st1[a])(H) → adP(a, st1[a])(H)) → ((adP ∧_ st1))(H) ⊦ a ∈ st0, adP(a, d)(H) ∧ ((adP ∧_ st0))(H)

User applied rule all left

And obtained the following new goals

Proof step 10:
adP0(a, d)(H), ((adP0 ∧_ st0))(H), a ∈ st0[a, d] ∧ adP0(a, st0[a, d][a])(H) → adP(a, st0[a, d][a])(H), ∀ a0, H. a0 ∈ st0[a, d] ∧ adP0(a0, st0[a, d][a0])(H) → adP(a0, st0[a, d][a0])(H), ∀ st1, adP0, H, adP. # st1 < # st0[a, d] → ((adP0 ∧_ st1))(H) ∧ (∀ a, H. a ∈ st1 ∧ adP0(a, st1[a])(H) → adP(a, st1[a])(H)) → ((adP ∧_ st1))(H) ⊦ a ∈ st0, adP(a, d)(H) ∧ ((adP ∧_ st0))(H)

Rule argument

pos: left 3 terms: a, H abort: F discard: F computed: T

Additional information

Used terms: a, H