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Proof step 19 from proof icrash-refsize-cons (previous step)

indexblock(istore1[n].indlist, istore1[n].ladrlist, istore0[n].ref-size) = indexblock(indx, lax, m), n ∈ istore0, ∀ n. n ∈ istore0 → istore0[n].ref-size = (as | n).size ⊦ m = (as | n).size

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
n ∈ istore0, ∀ n. n ∈ istore0 → istore0[n].ref-size = (as | n).size ⊦ istore0[n].ref-size = (as | n).size

Rule argument

Additional information

Used simplifier rules: ⊦ indexblock(indx, lax, m) = indexblock(indx0, lax0, m0) ↔ indx = indx0 ∧ lax = lax0 ∧ m = m0; (ws)