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Proof step 15 from proof junkref-union-disjoint (previous step)

st1 .refs ∩ st .refs = ∅, r ∈ st1, junkrefs(str, r ' + refs0, (st1 ∪ st)[r -- _type] .type, st1 ∪ 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 14:
st1 .refs ∩ st .refs = ∅, r ∈ st1, junkrefs(str, r ' + refs0, (st1 ∪ st)[r -- _type] .type, st1 ∪ st) ⊦ r ' + refs0 ≠ [] ∧ (r ' + refs0) .first = r ∧ junkrefs(str, r ' + refs0, st1[(r ' + refs0) .first -- _type] .type, st1), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st1[refs .first -- _type] .type, st1)

Rule argument

[EXRARG : pos: right 1, [extquanttermlist: (r ' + refs0) abort:F , discard:F , computed:F ]]

Additional information

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