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Proof step 2 from proof except-refsizeeq-write-left (previous step)

st0 =○refsize st1 ⊦ st0[n, ablock] =○refsize st1 ↔ n ∈ st0 ∧ ablock.list = st0[n].list ∧ ablock.ladrlist = st0[n].ladrlist

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist ⊦ (∀ n0. (n0 ∈ st0[n, ablock] ↔ n0 ∈ st1) ∧ st0[n, ablock][n0].list = st1[n0].list ∧ st0[n, ablock][n0].ladrlist = st1[n0].ladrlist) ↔ n ∈ st0 ∧ ablock.list = st0[n].list ∧ ablock.ladrlist = st0[n].ladrlist

Rule argument

Additional information

Used simplifier rules: except-refsize-eq: ⊦ st0 =○refsize st1 ↔ (∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist);