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Proof step 17 from proof inv-allocate-gnd (previous step)

remove-refsize0(gstore0) = gstore0, remove-refsize0(istore1) = istore1, elems nx ∩ ginval = ∅, dom gstore0 ∪ elems nx = dom gstore ∪ ginval, inv(gstore \ ginval), inv(gstore0), inv(istore2), inv(istore1), (dom gstore).max + 1 ∈ gstore, istore1 ⊑ istore2, pstate.journal? → nx ≠ [] ∧ pstate.lnum = nx.last ∧ pstate.lnum ∈ gstore, pstate.index? → pstate.lnum ∈ istore2 ⊦ pstate = readonly, dups(nx)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ gstore ≠ ∅ ∧ n > (dom gstore).max → (n ∈ gstore ↔ false)

Rule argument

spec: nat-store+set, inst: group-store name: dom-max-in seq: ⊦ gstore ≠ ∅ ∧ n > (dom gstore).max → (n ∈ gstore ↔ false) subst: [gstore, n] → [gstore, (dom gstore).max + 1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m > n ↔ n < m; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ dom ∅ = ∅; (s)% ⊦ ns ∪ ∅ = ns; (s)%una: ⊦ (ns ∪ ns0) ∪ ns1 = ns ∪ ns0 ∪ ns1; from specification set-union[nat-set]%unc: ⊦ ns ∪ ns0 = ns0 ∪ ns; from specification set-union[nat-set]% ⊦ ∅ \ ns = ∅; (s)% ⊦ inv(∅); (s)% ⊦ ¬ n ∈ ∅; (s) Used terms: gstore, (dom gstore).max + 1