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Proof step 9 from proof del-refs-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
r -- sk ∈ st, # st ≤ # (st -- r), r ∈ st, st -- r ⊆ st -- (r -- sk) → # (st -- r) ≤ # st - 1 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ n -1 = n - 1], [CSIMPSEQ : ⊦ rk ∈ st → # (st -- rk) = (# st) -1])]]