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Proof step 18 from proof junkref-subset (previous step)

st[r -- _type] = st1[r -- _type], r -- _type ∈ st1, junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1), r ∈ st1, ∀ rk. rk ∈ st → rk ∈ st1 ∧ st[rk] = st1[rk] ⊦ junkrefs(str, r ' + refs1, st1[r -- _type] .type, st), ∃ 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 17:
st[r -- _type] = st1[r -- _type], r -- _type ∈ st1, junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1), r ∈ st1, ∀ rk. rk ∈ st → rk ∈ st1 ∧ st[rk] = st1[rk] ⊦ junkrefs(str, r ' + refs1, st1[r -- _type] .type, st), 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 2, [extquanttermlist: (r ' + refs1) abort:F , discard:F , computed:F ]]

Additional information

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