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Proof step 52 from proof subset-size (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 51:
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

Rule argument

[EXRARG : pos: left 5, [extquanttermlist: (st0 -- r)abort:F, discard:F, computed:F]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0 -- r)]]