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Proof step 27 from proof subset-size (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 26:
r ∈ st0, r -- sk ∈ st0, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), # st0 .refs < # st1 .refs, r ∈ st1 .refs, # st1 .refs ≤ # st0[r -- sk, st1[r -- sk]] .refs ∨ ¬ st1 ⊆ st0[r -- sk, st1[r -- sk]], ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st1 = ∅

Rule argument

[EXRARG : pos: left 6, [extquanttermlist: (st0[r -- sk, st1[r -- sk]]) abort:F , discard:F , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0[r -- sk, st1[r -- sk]])]]