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Proof step 48 from proof subset-size (previous step)

r ∈ st0, r -- sk ∈ st0, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), # st0 .refs ≤ # st1 .refs, r ∈ st0 .refs, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, st1 ⊆ st0 -- r, r ∈ st1 .refs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 37:
⊦ ¬ r ∈ st → (st ⊆ st0 -- r ↔ st ⊆ st0)
Proof step 38:
⊦ ¬ r ∈ st .refs → ¬ r ∈ st
Proof step 47:
¬ r ∈ st1, r ∈ st0, r -- sk ∈ st0, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), # st0 .refs ≤ # st1 .refs, r ∈ st0 .refs, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, st1 ⊆ st0, r ∈ st1 .refs

Rule argument

[speclemmaarg: ,¬ r ∈ st → (st ⊆ st0 -- r ↔ st ⊆ st0) ,[r, st0, st] → [r, st0, st1] ,F ,storefuns , ,subset-delete ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, st0, st1)]]