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Proof step 13 from proof except-refsizeeq-write-left (previous step)

st1[n] = mkblock(x0, lax0, m1), st0[n] = mkblock(x0, lax0, m0), st0[n, mkblock(x0, lax0, m)][n0] = mkblock(x, lax, m2), n ∈ st0, n ∈ st1, ∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist ⊦ (n0 ∈ st0[n, mkblock(x0, lax0, m)] ↔ n0 ∈ st1) ∧ x = st1[n0].list ∧ lax = st1[n0].ladrlist

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
st0[n, mkblock(x0, lax0, m)][n0] = mkblock(x, lax, m2), st0[n] = mkblock(x0, lax0, m0), st1[n] = mkblock(x0, lax0, m1), st1[n0] = mkblock(x1, lax1, m3), n ∈ st1, n ∈ st0, ∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist ⊦ (n0 ∈ st0[n, mkblock(x0, lax0, m)] ↔ n0 ∈ st1) ∧ x = x1 ∧ lax = lax1

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] → [st1[n0], x1, lax1, m3]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: st1[n0], x1, lax1, m3