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Proof step 56 from proof pathexplus-pathex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 55:
validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st) ⊦ refs .first = refs .last, refs = [], ∃ refs0. refs0 ≠ [] ∧ refs0 .first = refs .first ∧ validrefs(str, refs0, mkclasstype(`java.lang.Object' .class), st) ∧ step∃(str, refs0 .last, refs .last, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]