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Proof step 27 from proof junkref-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
junkrefs(str, refs, st1[refs .first -- _type] .type, st1), st ⊆ st1 ⊦ refs = [], ∃ refs0. refs0 ≠ [] ∧ refs0 .first = refs .first ∧ junkrefs(str, refs0, st[refs .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]