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Proof step 51 from proof subset-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
⊦ (st -- r) .refs = st .refs -- r
Proof step 33:
⊦ ¬ r ∈ st .refs → ¬ r ∈ st
Proof step 50:
# (st0 .refs -- r) < # 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

Rule argument

[emptyarg]

Additional information

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