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Proof step 3 from proof remove-refsize0-one (previous step)

⊦ ∀ n0. (n0 ∈ remove-refsize0((∅)[n, mkblock(x, lax, m0)]) ↔ n0 ∈ (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])) ∧ ( n0 ∈ remove-refsize0((∅)[n, mkblock(x, lax, m0)]) → remove-refsize0((∅)[n, mkblock(x, lax, m0)])[n0] = (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])[n0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (n0 ∈ remove-refsize0((∅)[n, mkblock(x, lax, m0)]) ↔ n0 ∈ (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])) ∧ ( n0 ∈ remove-refsize0((∅)[n, mkblock(x, lax, m0)]) → remove-refsize0((∅)[n, mkblock(x, lax, m0)])[n0] = (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])[n0])

Proof step 3 from proof remove-refsize0-one (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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