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Proof step 14 from proof ex-validrefs-fits (conclusion)

validrefs(str, refs, ty, st), refs ≠ [] ⊦ ∃ sks. fits(refs .first, sks, st) ∧ st[refs .first -- sks] .refval = refs .last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
validrefs(str, refs, ty, st) ⊦ refs = [], ∃ sks. fits(refs .first, sks, st) ∧ st[refs .first -- sks] .refval = refs .last

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]