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Proof step 9 from proof pathex-levelsubset (previous step)

⊦ level⊆(str, str0) ∧ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st)) → (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str0, refs, mkclasstype(`java.lang.Object' .class), st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
level⊆(str, str0), validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st) ⊦ refs = [], ∃ refs0. refs0 ≠ [] ∧ refs0 .first = refs .first ∧ refs0 .last = refs .last ∧ validrefs(str0, refs0, mkclasstype(`java.lang.Object' .class), st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]