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Proof step 17 from proof unchanged-junkref (previous step)

unchanged(str, r, st, st1), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st1[r -- _type] .type, st1) ⊦ junkref(str, r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
unchanged(str, refs .first, st, st1), junkrefs(str, refs, st1[refs .first -- _type] .type, st1) ⊦ refs = [], junkref(str, refs .first, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]