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Proof step 3 from proof except-refsizeeq-write-left (previous step)

∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist ⊦ (∀ n0. (n0 ∈ st0[n, ablock] ↔ n0 ∈ st1) ∧ st0[n, ablock][n0].list = st1[n0].list ∧ st0[n, ablock][n0].ladrlist = st1[n0].ladrlist) ↔ n ∈ st0 ∧ ablock.list = st0[n].list ∧ ablock.ladrlist = st0[n].ladrlist

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
ablock = mkblock(x, lax, m), ∀ n. (n ∈ st0 ↔ n ∈ st1) ∧ st0[n].list = st1[n].list ∧ st0[n].ladrlist = st1[n].ladrlist ⊦ (∀ n0. (n0 ∈ st0[n, ablock] ↔ n0 ∈ st1) ∧ st0[n, ablock][n0].list = st1[n0].list ∧ st0[n, ablock][n0].ladrlist = st1[n0].ladrlist) ↔ n ∈ st0 ∧ x = st0[n].list ∧ lax = st0[n].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] → [ablock, x, lax, m, m0, m1, boolvar]

Additional information

Used simplifier rules: ⊦ (x = ablock.list ∧ lax = ablock.ladrlist) ∧ m = ablock.ref-size ↔ ablock = mkblock(x, lax, m); (elim) Used terms: ablock, x, lax, m, m0, m1, boolvar