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Proof step 49 from proof subset-size (previous step)

r ∈ st0 .refs, # st0 .refs - 1 < # st1 .refs, # st0 .refs ≤ # st1 .refs, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), r -- sk ∈ st0, r ∈ st0, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ r ∈ st1 .refs, st1 ⊆ st0 -- r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
⊦ st .refs = ∅ ↔ st = ∅
Proof step 48:
r ∈ st0, r -- sk ∈ st0, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), # st0 .refs ≤ # st1 .refs, r ∈ st0 .refs, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, st1 ⊆ st0 -- r, r ∈ st1 .refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m ≤ n → (n < m ↔ false)], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0], [CSIMPSEQ : ⊦ ¬ m ≤ n ↔ n < m], [CSIMPSEQ : ⊦ m < 1 ↔ m = 0], [CSIMPFORWARD : ⊦ # rset = 0 → rset = ∅], [CSIMPFORWARD : ⊦ r ∈ rset → rset ≠ ∅])]]