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Proof step 10 from proof refsize-cons-write (previous step)

st[n] = mkblock(x0, lax0, m0), st[n, mkblock(x, lax, m)][n0] = mkblock(x1, lax1, m1), n0 ∈ st[n, mkblock(x, lax, m)], ¬ n ∈ st ∨ m0 = (as | n).size → m = (as | n).size, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m1 = (as | n0).size

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 11:
st[n, mkblock(x, lax, m)][n0] = mkblock(x1, lax1, m1), st[n] = mkblock(x0, lax0, m0), n0 = n, n0 ∈ st[n, mkblock(x, lax, m)], ¬ n ∈ st ∨ m0 = (as | n).size → m = (as | n).size, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m1 = (as | n0).size
Proof step 13:
st[n, mkblock(x, lax, m)][n0] = mkblock(x1, lax1, m1), st[n] = mkblock(x0, lax0, m0), n0 ∈ st[n, mkblock(x, lax, m)], ¬ n ∈ st ∨ m0 = (as | n).size → m = (as | n).size, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ m1 = (as | n0).size, n0 = n

Rule argument

fma: n0 = n

Additional information

Used formulas: n0 = n