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Proof step 33 from proof refs-equal (previous step)

# st .refs = # st0 .refs, st ⊆ st0, r ∈ st0, r0 ∈ st .refs, r0 ∈ st0 .refs, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ st = ∅, st0 = ∅, r ∈ st, r ∈ st0 -- r0 ∧ st -- r0 ⊆ st0 -- r0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ r ≠ r1 → (r ∈ st -- r1 ↔ r ∈ st)
Proof step 19:
⊦ ¬ r ∈ st → ¬ r ∈ st .refs
Proof step 32:
r ≠ r0, # st .refs = # st0 .refs, st ⊆ st0, r ∈ st0, r0 ∈ st .refs, r0 ∈ st0 .refs, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ st = ∅, st0 = ∅, r ∈ st, r ∈ st0 ∧ st -- r0 ⊆ st0 -- r0

Rule argument

[speclemmaarg: ,r ≠ r1 → (r ∈ st -- r1 ↔ r ∈ st) ,[r1, st, r] → [r0, st0, r] ,F ,storefuns , ,refin-del ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r0, st0, r)]]