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Proof step 7 from proof pathex-levelsubset (previous step)

refs = r + refs1, level⊆(str, str0), validrefs(str, refs, mkclasstype(`java.lang.Object' .class), st) ⊦ refs = [], ∃ refs0. refs0 ≠ [] ∧ refs0 .first = r ∧ refs0 .last = refs .last ∧ validrefs(str0, refs0, mkclasstype(`java.lang.Object' .class), st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
level⊆(str, str0), validrefs(str, r ' + refs1, mkclasstype(`java.lang.Object' .class), st) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = (refs .first ' + refs1) .last ∧ validrefs(str0, refs, mkclasstype(`java.lang.Object' .class), st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : list, references, e-01, ⊦ r ' + refs ≠ []], [CSIMPNAMEDSEQ : list, references, c, ⊦ r + refs = r ' + refs])]]