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Proof step 10 from proof pathex-levelsubset-01 (previous step)

level⊆(str, str0), validref(str, r, ty, st), validrefs(str0, r ' + refs0, st[r -- _type] .type, st) ⊦ refs0 = [], r = refs0 .last, dups(refs0), r ∈ refs0, ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = refs0 .last ∧ validrefs(str, refs, st[refs .first -- _type] .type, st)

User applied rule exists right

And obtained the following new goals

Proof step 9:
level⊆(str, str0), validref(str, r, ty, st), validrefs(str0, r ' + refs0, st[r -- _type] .type, st) ⊦ refs0 = [], r = refs0 .last, dups(refs0), r ∈ refs0, r ' + refs0 ≠ [] ∧ (r ' + refs0) .first = r ∧ (r ' + refs0) .last = refs0 .last ∧ validrefs(str, r ' + refs0, st[(r ' + refs0) .first -- _type] .type, st)

Rule argument

[EXRARG : pos: right 5, [extquanttermlist: (r ' + refs0) abort:F , discard:T , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r ' + refs0)]]