Proof step 1 from proof step-absstep-I1-I2-tau-same (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ls0.sz = # gs0.ar, ls.tid = als.tid, lsf(tid).tid = tid, alsf(tid).tid = tid, I2 = pcstepf(gs, ls, I1, τ), alsf(tid) = als, apc = invInsert, apcf(tid) = apc, gs0 = gstepf(gs, ls, I1, τ), ls0 = lstepf(gs, ls, I1, τ), lsf(tid) = ls, lsf0(tid) = ls0, pcf(tid) = I1, pcf0(tid) = I2, # gs.ar ≠ 0, # gs0.ar ≠ 0, ls.e ≠ ⊥, ls0.e ≠ ⊥, SetLInv(ags, als, apc), absset(gs.ar, ags.set), htok(gs0.ar), htok(gs.ar), pre(gs, ls, I1, τ), apc = invMember ∨ apc = invInsert → ls.e = als.le, ∀ tid0. tid0 ≠ tid → HashsetLInv(gs, lsf(tid0), pcf(tid0)) ∧ SetLInv(ags, alsf(tid0), apcf(tid0)) ∧ labs(gs, lsf(tid0), pcf(tid0), ags, alsf(tid0), apcf(tid0)), ∀ tid0. tid0 ≠ tid → alsf(tid0).tid = tid0, ∀ tid0. tid0 ≠ tid → lsf(tid0).tid = tid0 ⊦ absset(gs0.ar, ags.set) ∧ ls0.tid = als.tid ∧ apc = invInsert ∧ (apc = invMember ∨ apc = invInsert → ls0.e = als.le)

Rule argument

Additional information

Proof step 2 from proof step-absstep-I1-I2-tau-same (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  lsf(tid).e = alsf(tid).le, lsf0(tid).sz = # gs.ar, lsf(tid).tid = tid, alsf(tid).tid = tid, apcf(tid) = invInsert, lsf0(tid) = lsf(tid).sz:= # gs.ar, pcf(tid) = I1, pcf0(tid) = I2, # gs.ar ≠ 0, lsf0(tid).e ≠ ⊥, alsf(tid).le ≠ ⊥, SetLInv(ags, alsf(tid), invInsert), absset(gs.ar, ags.set), htok(gs.ar), ∀ tid0. tid0 ≠ tid → HashsetLInv(gs, lsf(tid0), pcf(tid0)) ∧ SetLInv(ags, alsf(tid0), apcf(tid0)) ∧ labs(gs, lsf(tid0), pcf(tid0), ags, alsf(tid0), apcf(tid0)), ∀ tid0. alsf(tid0).tid = tid0, ∀ tid0. lsf(tid0).tid = tid0 ⊦ lsf0(tid).tid = tid ∧ lsf0(tid).e = alsf(tid).le

Rule argument

Additional information

Used simplifier rules: ⊦ gstepf(gs, ls, I1, τ) = gs; (s) ⊦ pre(gs, ls, I1, a) ↔ a = τ; (s) ⊦ pcstepf(gs, ls, I1, a) = I2; (s) ⊦ lstepf(gs, ls, I1, τ) = ls.sz:= # gs.ar; (s)

Proof step 3 from proof step-absstep-I1-I2-tau-same (previous step) (back to tree)

Heuristic use patterns applied rule apply elim lemma

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkls(tid, e, b, sz, sz0, sz1, e0).sz:= sz2 = mkls(tid, e, b, sz2, sz0, sz1, e0); (s) ⊦ mkls(tid, e, b, sz, sz0, sz1, e0).sz = sz; (s) ⊦ mkls(tid, e, b, sz, sz0, sz1, e0).tid = tid; (s) ⊦ mkls(tid, e, b, sz, sz0, sz1, e0).e = e; (s) ⊦ (((((tid = ls.tid ∧ e = ls.e) ∧ (b ↔ ls.b)) ∧ sz = ls.sz) ∧ sz0 = ls.n0) ∧ n = ls.n) ∧ e0 = ls.e0 ↔ ls = mkls(tid, e, b, sz, sz0, n, e0); (elim) Used terms: lsf(tid), b, tid1, n0, e0, e, n, n1