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Proof step 5 from proof junkref-levelsubset (previous step)

level⊆(str, str0), junkrefs(str0, r ' + refs0, st[r -- _type] .type, st) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
level⊆(str, str0), junkrefs(str0, r ' + refs0, st[r -- _type] .type, st) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]