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Proof step 29 from proof newrefs-two (conclusion)

⊦ ¬ (∃ r, r0. refs = r + r0 ∧ newref(r, st) ∧ newref(r0, st) ∧ r ≠ r0) → newref_list(2, st) ≠ refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
⊦ ∃ r, r0. newref_list(2, st) = r ' + r0 ' ∧ newref(r, st) ∧ newref(r0, st) ∧ r ≠ r0

Rule argument

[emptyarg]

Additional information

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