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Proof step 3 from proof at-explicit-other (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
st[m] = mkblock(x, lax, m2), @(m0, m1) ∈ lax, m ∈ st ⊦ st[n, ablock][@(m, m0, m1)] = x[pos(@(m0, m1), lax)], m = n

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[m], x, lax, m2, m3, m4, boolvar]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st[m], x, lax, m2, m3, m4, boolvar