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Proof step 5 from proof at-write-same (previous step)

n ∈ st, m0 ∈ st, @(m1, m2) ∈ st[m0].ladrlist ⊦ st[n, mkblock(st[n].list, st[n].ladrlist, m)][@(m0, m1, m2)] = st[m0].list[pos(@(m1, m2), st[m0].ladrlist)]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
st[n] = mkblock(x, lax, m3), @(m1, m2) ∈ st[m0].ladrlist, m0 ∈ st, n ∈ st ⊦ st[n, mkblock(x, lax, m)][@(m0, m1, m2)] = st[m0].list[pos(@(m1, m2), st[m0].ladrlist)]

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, m0, m1, boolvar] → [st[n], x, lax, m5, m6, m7, boolvar0]

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], x, lax, m5, m6, m7, boolvar0