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Proof step 7 from proof validrefs-append-01 (conclusion)

⊦ validrefs(str, refs, ty, st) ∧ validrefs(str, refs0, ty0, st) ∧ refs .last = refs0 .first ∧ refs ≠ [] ∧ refs0 ≠ [] → validrefs(str, refs + refs0 .rest, ty, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
refs .last = refs0 .first, validrefs(str, refs0, ty0, st), validrefs(str, refs, ty, st) ⊦ refs = [], refs0 = [], validrefs(str, refs + refs0 .rest, ty, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]