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Proof step 3 from proof sw-del-add-01 (conclusion)

r ≠ null ∧ ¬ r ∈ H ⊦ r0 ∈ H1 ∧ ¬ r ∈ H1 → ((H -- r0) ++ r)[r, o] = (H ++ r)[r, o] -- r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
r0 ∈ H1 ⊦ r = null, ((H -- r0) ++ r)[r, o] = (H ++ r)[r, o] -- r0, r ∈ H, r ∈ H1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]