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Proof step 6 from proof unchanged-validrefs-unchNorec (previous step)

refs = r + refs0 ∧ refs ≠ [], r0 ∈ refs, validrefs(str, refs, ty, st), unchanged(str, r, st, st0) ⊦ unchangedNorec(r0, st, st0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
unchanged(str, r, st, st0), validrefs(str, r ' + refs0, ty, st) ⊦ unchangedNorec(r0, st, st0), r0 ≠ r ∧ ¬ r0 ∈ refs0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]