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Proof step 7 from proof remove-refsize0-one (previous step)

⊦ (n0 ∈ (∅)[n, mkblock(x, lax, m0)] ∧ (∅)[n, mkblock(x, lax, m0)][n0].ref-size ≠ 0 ↔ n0 ∈ (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])) ∧ ( n0 ∈ (∅)[n, mkblock(x, lax, m0)] ∧ (∅)[n, mkblock(x, lax, m0)][n0].ref-size ≠ 0 → (∅)[n, mkblock(x, lax, m0)][n0] = (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])[n0])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
(∅)[n, mkblock(x, lax, m0)][n0] = mkblock(x0, lax0, m) ⊦ (n0 ∈ (∅)[n, mkblock(x, lax, m0)] ∧ m ≠ 0 ↔ n0 ∈ (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])) ∧ (n0 ∈ (∅)[n, mkblock(x, lax, m0)] ∧ m ≠ 0 → (∅)[n, mkblock(x, lax, m0)][n0] = (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])[n0])

Rule argument

spec: ablock name: elim seq: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m) subst: [ablock, x, lax, m] → [(∅)[n, mkblock(x, lax, m0)][n0], x0, lax0, m]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: (∅)[n, mkblock(x, lax, m0)][n0], x0, lax0, m