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Proof step 9 from proof pathex-subset-equiv (previous step)

st ⊆ st0, validrefs(str, r ' + refs0, mkclasstype(`java.lang.Object' .class), st0) ⊦ refs0 = [], junkref(str, r, st), r ' + refs0 ≠ [] ∧ (r ' + refs0) .first = r ∧ (r ' + refs0) .last = refs0 .last ∧ validrefs(str, r ' + refs0, mkclasstype(`java.lang.Object' .class), st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
st ⊆ st0, validrefs(str, r ' + refs0, mkclasstype(`java.lang.Object' .class), st0) ⊦ refs0 = [], junkref(str, r, st), validrefs(str, r ' + refs0, mkclasstype(`java.lang.Object' .class), st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ refs ≠ [] → (r ' + refs) .last = refs .last], [CSIMPSEQ : ⊦ (r ' + refs) .first = r], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' ≠ []], [CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []])]]