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Proof step 58 from proof pathexplus-pathex (previous step)

path∃(str, r, r0, st) ⊦ path∃+(str, r, r0, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ path∃+(str, r, r0, st) ↔ r = r0 ∨ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st) ∧ step∃(str, refs .last, r0, st))
Proof step 57:
path∃(str, r, r0, st) ⊦ r = r0 ∨ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st) ∧ step∃(str, refs .last, r0, st))

Rule argument

[crewrite-arg: pathex-plus-def , F , [st, r0, r, str] → [st, r0, r, str] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r0, r, str)]]