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Proof step 8 from proof del-refs-size (previous step)

r -- sk ∈ st, # st ≤ # (st -- r), r ∈ st, st -- r ⊆ st -- (r -- sk) → # (st -- r) ≤ # st - 1 ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
⊦ ¬ r ∈ ∅
Proof step 7:
# st = m + 1 ∧ # st ≠ 0, r -- sk ∈ st, # st ≤ # (st -- r), r ∈ st, st -- r ⊆ st -- (r -- sk) → # (st -- r) ≤ m ⊦

Rule argument

[speclemmaarg: n ≠ 0 ,m = n - 1 ↔ n = m + 1 ,[n, m] → [# st, m] ,T ,nat , ,elim-pred-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # st = 0 ↔ st = ∅], [CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (# st, m)]]