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Proof step 3 from proof Sep-any (previous step)

((adP ⊗_ st * P))(H) ⊦ st = ∅, ∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * P))(H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0)
Proof step 5:
((adP ⊗_ st * P))(H) ⊦ ∀ a, d, st0. ¬(st = st0[a, d] ∧ ¬ a ∈ st0), ∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * P))(H)

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