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Proof step 9 from proof pathex-levelsubset-01 (previous step)

level⊆(str, str0), validref(str, r, ty, st), validrefs(str0, r ' + refs0, st[r -- _type] .type, st) ⊦ refs0 = [], r = refs0 .last, dups(refs0), r ∈ refs0, r ' + refs0 ≠ [] ∧ (r ' + refs0) .first = r ∧ (r ' + refs0) .last = refs0 .last ∧ validrefs(str, r ' + refs0, st[(r ' + refs0) .first -- _type] .type, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
level⊆(str, str0), validref(str, r, ty, st), validrefs(str0, r ' + refs0, st[r -- _type] .type, st) ⊦ refs0 = [], r = refs0 .last, dups(refs0), r ∈ refs0, validrefs(str, r ' + refs0, st[r -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list-last, references, l-02, ⊦ refs ≠ [] → (r ' + refs) .last = refs .last], [CSIMPNAMEDSEQ : list, references, c-08, ⊦ (r ' + refs) .first = r], [CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : list, references, e, ⊦ r ' ≠ []], [CSIMPNAMEDSEQ : list, references, app-02-ant, ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []])]]