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Proof step 16 from proof nodes-alloc-append (previous step)

⊦ @(m2, m3, m4) = @(n, m, m0), n ∈ st, (m2 ∈ st[n, mkblock(a ', @(m, m0) ', m1)] ∧ @(m3, m4) ∈ st[n, mkblock(a ', @(m, m0) ', m1)][m2].ladrlist → @(m2, m3, m4) ∈ st) ∧ ( m2 ∈ st[n, mkblock(a ', @(m, m0) ', m1)] ∧ @(m3, m4) ∈ st[n, mkblock(a ', @(m, m0) ', m1)][m2].ladrlist → st[n, mkblock(a ', @(m, m0) ', m1)][@(m2, m3, m4)] = nodes(st)[@(m2, m3, m4)])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
st[n, mkblock(a ', @(m, m0) ', m1)][m2] = mkblock(x, lax, m5) ⊦ @(m2, m3, m4) = @(n, m, m0), n ∈ st, (m2 ∈ st[n, mkblock(a ', @(m, m0) ', m1)] ∧ @(m3, m4) ∈ lax → @(m2, m3, m4) ∈ st) ∧ (m2 ∈ st[n, mkblock(a ', @(m, m0) ', m1)] ∧ @(m3, m4) ∈ lax → st[n, mkblock(a ', @(m, m0) ', m1)][@(m2, m3, m4)] = nodes(st)[@(m2, m3, m4)])

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(a ', @(m, m0) ', m1)][m2], x, lax, 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(a ', @(m, m0) ', m1)][m2], x, lax, m5