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Proof step 2 from proof except-refsizeeq-trans (previous step)

st0 =○refsize st1, st1 =○refsize st2 ⊦ st0 =○refsize st2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n. (n ∈ st1 ↔ n ∈ st2) ∧ st1[n].list = st2[n].list ∧ st1[n].ladrlist = st2[n].ladrlist, ∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist ⊦ (n ∈ st0 ↔ n ∈ st2) ∧ st0[n].list = st2[n].list ∧ st0[n].ladrlist = st2[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);