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Proof step 11 from proof unchanged-junkref-level (previous step)

level⊆(str, str0), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st), unchanged(str0, r, st, st1) ⊦ junkref(str, r, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
level⊆(str, str0), junkrefs(str, refs, st[refs .first -- _type] .type, st), unchanged(str0, refs .first, st, st1) ⊦ refs = [], junkref(str, refs .first, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]