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Proof step 64 from proof newrefs-three (conclusion)

⊦ ¬ (∃ r, r0, r1. refs = r + (r0 + r1) ∧ newref(r1, st) ∧ newref(r, st) ∧ newref(r0, st) ∧ r ≠ r0 ∧ r ≠ r1 ∧ r0 ≠ r1) → newref_list(3, st) ≠ refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 63:
⊦ ∃ r, r0, r1. newref_list(3, st) = r ' + r0 ' + r1 ' ∧ newref(r1, st) ∧ newref(r, st) ∧ newref(r0, st) ∧ r ≠ r0 ∧ r ≠ r1 ∧ r0 ≠ r1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r + refs = r ' + refs], [CSIMPSEQ : ⊦ r + r0 = r ' + r0 '])]]