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Proof step 4 from proof pathex-lookup-put-other (conclusion)

⊦ ¬ path∃(str, r0, r1, st) ∧ is_valid_ref(str, r1, st) → (st[r1 -- sk, val][r0 -- sk0]) = st[r0 -- sk0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
is_valid_ref(str, r1, st) ⊦ (st[r1 -- sk, val][r0 -- sk0]) = st[r0 -- sk0], path∃(str, r0, r1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]