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Proof step 21 from proof reffield-del (previous step)

¬ ¬ is_valid_ref(r0, st -- r), has_reffield(r0, st -- r) ⊦ has_reffield(r0, st), r0 = r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ r ≠ r0 → (is_valid_ref(r, st -- r0) ↔ is_valid_ref(r, st))
Proof step 20:
has_reffield(r0, st -- r), is_valid_ref(r0, st) ⊦ r0 = r, has_reffield(r0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]