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Proof step 22 from proof subset-size (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ st ⊆ st0[rk, val] ↔ st ⊆ st0 ∧ ¬ rk ∈ st ∨ rk × val ∈ st ∧ st -- rk ⊆ st0 -- rk
Proof step 21:
r ∈ st1 .refs, # st0 .refs < # st1 .refs, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), r -- sk ∈ st0, r ∈ st0, r ∈ st0 .refs, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st1 = ∅, st0 = ∅, st1 ⊆ st0 ∧ ¬ r -- sk ∈ st1 ∨ (r -- sk) × st1[r -- sk] ∈ st1 ∧ st1 -- (r -- sk) ⊆ st0 -- (r -- sk)

Rule argument

[speclemmaarg: , st ⊆ st0[rk, val] ↔ st ⊆ st0 ∧ ¬ rk ∈ st ∨ rk × val ∈ st ∧ st -- rk ⊆ st0 -- rk ,[val, rk, st0, st] → [st1[r -- sk], r -- sk, st0, st1] ,F ,store ,javastore ,subset-put-01 ]

Additional information

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