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Proof step 12 from proof junkref-subset (previous step)

st[r -- _type] = st1[r -- _type], r ∈ st1, junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1), r -- _type ∈ st1, rk ∈ st, ∀ 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), rk ∈ st1 ∧ st[rk] = st1[rk]

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 6, [extquanttermlist: (rk)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])], [EXTRATERMS : (rk)]]