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Proof step 10 from proof remove-refsize0-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(∅)[n0] = mkblock(x0, lax0, m), n0 ∈ (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)]) ⊦ n0 = n

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n ∈ ∅; (s)% ⊦ n ≠ n0 → (n ∈ st[n0, ablock] ↔ n ∈ st); (s)% ⊦ n ≠ n0 → st[n0, ablock][n] = st[n]; (s)