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Proof step 3 from proof size-del (previous step)

r ∈ H ⊦ #(H -- r) = # H - 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ r ≠ null → (r ∈ H ↔ (∃ H0, ce. H = (H0 ++ r)[r, ce] ∧ ¬ r ∈ H0))

Rule argument

spec: heap, inst: cell-heap name: exin seq: ⊦ r ≠ null → (r ∈ H ↔ (∃ H0, ce. H = (H0 ++ r)[r, ce] ∧ ¬ r ∈ H0)) subst: [H, r] → [H, r] flags: <all paths>

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)%r ∈ H ⊦ H[r, ce] -- r = H -- r; (s)%¬ r ∈ H, r ≠ null ⊦ (H ++ r) -- r = H; (s)%r ∈ H ⊦ # H[r, ce] = # H; (s)%¬ r ∈ H, r ≠ null ⊦ #(H ++ r) = # H + 1; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ¬ null ∈ H; (s) Used terms: H, r