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Proof step 3 from proof at-diff (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

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