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Proof step 4 from proof refsize-cons-remove-index-0 (previous step)

st[m] = mkblock(x, lax, n0), st[m, mkblock(x, lax, n0 - m1)][n] = mkblock(x0, lax0, m2), m ∈ st, @(m, m0, m1) ∈ as, n ∈ st[m, mkblock(x, lax, n0 - m1)], ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, m2 = ((as -- @(m, m0, m1)) | n).size

User applied rule all left

And obtained the following new goals

Proof step 5:
st[m, mkblock(x, lax, n0 - m1)][n] = mkblock(x0, lax0, m2), st[m] = mkblock(x, lax, n0), n ∈ st[m, mkblock(x, lax, n0 - m1)], @(m, m0, m1) ∈ as, m ∈ st, n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, m2 = ((as -- @(m, m0, m1)) | n).size

Rule argument

pos: left 6 terms: n abort: F discard: T computed: F

Additional information

Used terms: n