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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0)
Proof step 8:
((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), ∀ a, d, st0. ¬(st = st0[a, d] ∧ ¬ a ∈ st0)

Rule argument

spec: store name: exnotempty seq: ⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0) subst: [st] → [st] flags: <all paths>

Additional information

Used terms: st