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Proof step 13 from proof diff-in (previous step)

r ∈ (H0 -- r0) \ H ↔ r ≠ r0 ∧ r ∈ H0 ∧ ¬ r ∈ H, ∀ H2, r, H0. # H2 < # (H ++ r0)[r0, o] → true → (r ∈ H0 \ H2 ↔ r ∈ H0 ∧ ¬ r ∈ H2) ⊦ r ∈ (H0 -- r0) \ H ↔ r ∈ H0 ∧ r ≠ r0 ∧ ¬ r ∈ H, r0 = null, r0 ∈ H

User applied rule apply induction

And closed the goal!

Rule argument

subst: [H2, r, H0] → [H, r, H0]

Additional information

Used simplifier rules: in-alloc: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H;%size-write: r ∈ H ⊦ # H[r, o] = # H;%size-alloc: ¬ r ∈ H, r ≠ null ⊦ #(H ++ r) = # H + 1;% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: H, r, H0