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Proof step 10 from proof at-write-same (previous step)

st[m0] = mkblock(x0, lax0, m4), st[n] = mkblock(x, lax, m3), m0 ∈ st, n ∈ st, @(m1, m2) ∈ lax0 ⊦ st[n, mkblock(x, lax, m)][m0].list[pos(@(m1, m2), st[n, mkblock(x, lax, m)][m0].ladrlist)] = x0[pos(@(m1, m2), lax0)]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
st[n] = mkblock(x, lax, m3), st[m0] = mkblock(x0, lax0, m4), st[n, mkblock(x, lax, m)][m0] = mkblock(x1, lax1, m5), @(m1, m2) ∈ lax0, n ∈ st, m0 ∈ st ⊦ x1[pos(@(m1, m2), lax1)] = x0[pos(@(m1, m2), lax0)]

Rule argument

spec: ablock name: elim seq: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m) subst: [ablock, x, lax, m] → [st[n, mkblock(x, lax, m)][m0], x1, lax1, m5]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st[n, mkblock(x, lax, m)][m0], x1, lax1, m5