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Proof step 6 from proof unchanged-junkref-level (previous step)

unchanged(str0, r, st, st1), level⊆(str, str0), junkrefs(str, r ' + refs0, st[r -- _type] .type, st) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st1[refs .first -- _type] .type, st1)

User applied rule exists right

And obtained the following new goals

Proof step 5:
unchanged(str0, r, st, st1), level⊆(str, str0), junkrefs(str, r ' + refs0, st[r -- _type] .type, st) ⊦ r ' + refs0 ≠ [] ∧ (r ' + refs0) .first = r ∧ junkrefs(str, r ' + refs0, st1[(r ' + refs0) .first -- _type] .type, st1)

Rule argument

[EXRARG : pos: right 1, [extquanttermlist: (r ' + refs0) abort:F , discard:T , computed:F ]]

Additional information

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