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Proof step 24 from proof junkref-subset (previous step)

st ⊆ st1, junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)

User applied rule exists right

And obtained the following new goals

Proof step 23:
st ⊆ st1, junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1) ⊦ r ' + refs1 ≠ [] ∧ (r ' + refs1) .first = r ∧ junkrefs(str, r ' + refs1, st[(r ' + refs1) .first -- _type] .type, st), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)

Rule argument

[EXRARG : pos: right 1, [extquanttermlist: (r ' + refs1) abort:F , discard:F , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r ' + refs1)]]