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Proof step 2 from proof refsize-cons-write (previous step)

refsize-cons(st, as) ⊦ refsize-cons(st[n, ablock], as) ↔ ¬ n ∈ st ∨ st[n].ref-size = (as | n).size → ablock.ref-size = (as | n).size

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ (∀ n0. n0 ∈ st[n, ablock] → st[n, ablock][n0].ref-size = (as | n0).size) ↔ ¬ n ∈ st ∨ st[n].ref-size = (as | n).size → ablock.ref-size = (as | n).size

Rule argument

Additional information

Used simplifier rules: refsize-cons-def: ⊦ refsize-cons(st, as) ↔ (∀ n. n ∈ st → st[n].ref-size = (as | n).size);