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Proof step 21 from proof subset-size (previous step)

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) × st1[r -- sk] ∈ st1 ∧ st1 -- (r -- sk) ⊆ st0 -- (r -- sk)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
r -- sk ∈ st0, r ∈ st0, r ∈ st0 .refs, r ∈ st1 .refs, # st0 .refs < # st1 .refs, st1 ⊆ st0 -- (r -- sk), ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st1 = ∅, st0 = ∅, r -- sk ∈ st1, st1 ⊆ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ rk ∈ st → st -- rk = st], [CSIMPSEQ : ⊦ rk × val ∈ st ↔ ¬ ¬ (rk ∈ st ∧ st[rk] = val)])]]