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Proof step 47 from proof subset-size (previous step)

¬ r ∈ st1, 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 ∈ st1 .refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
⊦ ¬ r ∈ st → ¬ r ∈ st .refs
Proof step 46:
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 ∈ st1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]