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Proof step 4 from proof remove-refsize0-one (previous step)

⊦ (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])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n ∈ remove-refsize0(st) ↔ n ∈ st ∧ st[n].ref-size ≠ 0
Proof step 6:
⊦ (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 → remove-refsize0((∅)[n, mkblock(x, lax, m0)])[n0] = (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])[n0])

Rule argument

spec: <toplevel> name: in-remove-refsize0 seq: ⊦ n ∈ remove-refsize0(st) ↔ n ∈ st ∧ st[n].ref-size ≠ 0 subst: [st, n] → [(∅)[n, mkblock(x, lax, m0)], n0] flags: <all paths>

Additional information

Used terms: (∅)[n, mkblock(x, lax, m0)], n0