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Proof step 5 from proof And-discard (previous step)

∀ st0, P. # st0 < # st → true → st0 = ∅ ∨ (λ a, d. P) ∧_ st0 = P ⊦ st = ∅, (λ a, d. P) ∧_ st = P

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0)
Proof step 7:
∀ st0, P. # st0 < # st → true → st0 = ∅ ∨ (λ a, d. P) ∧_ st0 = P ⊦ (λ a, d. P) ∧_ st = P, ∀ 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