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Proof step 66 from proof hasSharing-pathex (previous step)

∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st), hasSharing(r0, st) ⊦ hasSharing(r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
validrefs(str, refs, st[refs .first -- _type] .type, st), hasSharing(refs .last, st) ⊦ refs = [], hasSharing(refs .first, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]