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Proof step 7 from proof map-putex (previous step)

((r => o1 * P))((H0 ++ r)[r, o1]) ⊦ r = null, ((r => o * P))((H0 ++ r)[r, o]), r ∈ H0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ ((r => o * P))(H) ↔ P(H -- r) ∧ r ∈ H ∧ H[r] = o
Proof step 9:
((r => o1 * P))((H0 ++ r)[r, o1]) ⊦ r = null, r ∈ H0, P((H0 ++ r)[r, o] -- r) ∧ r ∈ (H0 ++ r)[r, o] ∧ (H0 ++ r)[r, o][r] = o

Rule argument

spec: <toplevel> name: map-del seq: ⊦ ((r => o * P))(H) ↔ P(H -- r) ∧ r ∈ H ∧ H[r] = o subst: [H, P, o, r] → [(H0 ++ r)[r, o], P, o, r] flags: <all paths>

Additional information

Used terms: (H0 ++ r)[r, o], P, o, r