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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
r ≠ null ∧ ¬ r ∈ H ⊦ r0 ≠ r → ((H -- r0) ++ r)[r, o] = (H ++ r)[r, o] -- r0

Rule argument

spec: <toplevel> name: sw-del-add seq: r ≠ null ∧ ¬ r ∈ H ⊦ r0 ≠ r → ((H -- r0) ++ r)[r, o] = (H ++ r)[r, o] -- r0 subst: [o, r, r0, H] → [o, r, r0, H] flags: <all paths>

Additional information

Used terms: o, r, r0, H