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Proof step 23 from proof subset-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ st .refs = ∅ ↔ st = ∅
Proof step 22:
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]]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r ∈ rset → rset ≠ ∅], [CSIMPSEQ : ⊦ m < n → (n ≤ m ↔ false)], [CSIMPSEQ : ⊦ r ∈ rset → rset ++ r = rset])]]