[Imprint][Data Privacy Statement][E-Mail]



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Proof step 1 from proof allwritesOK-ind-at-most-one-write (conclusion) (back to tree)

⊦ allWritesOKInd(pending, persisted, h) → #((h |W | t) | false) ≤ 1

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

allWritesOKInd(pending, persisted, h), ¬ #((h |W | t) | false) ≤ 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: restrict-W: ⊦ h |W = h.filter(λ vpe. vpe._1.W?); restrict-thread: ⊦ h | t = h.filter(λ vpe. vpe._1.t = t); restrict-vp: ⊦ h | b = h.filter(λ vpe. vpe._2 ↔ b); ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ # x.filter(p) = x.count(p); (s)

Proof step 3 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

h.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, h.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, allWritesOKInd(pending, persisted, h) ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

term: h

Additional information

Used simplifier rules: List('a) freely generated by [], +;; ⊦ [].filter(p) = []; (s) ⊦ [].count(p) = 0; (s)

Proof step 4 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

(vpe + h0).filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, (vpe + h0).filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, allWritesOKInd(pending, persisted, vpe + h0), ∀ t, pending, persisted. ¬ ( h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0 ∧ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1 ∧ allWritesOKInd(pending, persisted, h0)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ a '.filter(p) = (p(a) ⊃ a ';[]); (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n + 0 = n; (s) ⊦ [].count(p) = 0; (s) ⊦ [].filter(p) = []; (s) ⊦ a + x = a ' + x; (s) ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ (x + y).count(p) = x.count(p) + y.count(p); (s) ⊦ (x + y).filter(p) = x.filter(p) + y.filter(p); (s) ⊦ (a ' + x).filter(p) = (p(a) ⊃ a ';[]) + x.filter(p); (s)

Proof step 5 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

(vpe._1.W? ⊃ vpe ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending, persisted, vpe ' + h0), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ (vpe._1.W? ⊃ vpe ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 1, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ (vpe._1.t = t ⊃ vpe ';[]).count(λ vpe. ¬ vpe._2) = 1

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

term: vpe._1

Additional information

Used terms: e, boolvar

Proof step 6 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

vpe = (e, boolvar), (e.W? ⊃ vpe ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending, persisted, vpe ' + h0), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ (e.t = t ⊃ vpe ';[]).count(λ vpe. ¬ vpe._2) = 1, (e.W? ⊃ vpe ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 7 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

(e.W? ⊃ (e, boolvar) ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending, persisted, (e, boolvar) ' + h0), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ (e.W? ⊃ (e, boolvar) ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 1, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ (e.t = t ⊃ (e, boolvar) ';[]).count(λ vpe. ¬ vpe._2) = 1

User applied rule constructor cut

And obtained the following new goals

Rule argument

var: e

Additional information

Used simplifier rules: RWevent freely generated by R, W;;

Proof step 8 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

e = R(t1, l, v), (e.W? ⊃ (e, boolvar) ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending, persisted, (e, boolvar) ' + h0), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ (e.t = t ⊃ (e, boolvar) ';[]).count(λ vpe. ¬ vpe._2) = 1, (e.W? ⊃ (e, boolvar) ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: allWritesOKInd-recR: ⊦ allWritesOKInd(pending, persisted, (R(t, l, v), b) ' + h) ↔ allWritesOKInd(pending, persisted, h); ⊦ [].count(p) = 0; (s) ⊦ [].filter(p) = []; (s) ⊦ ¬ R(t, l, v).W?; (s) a: ⊦ (x + y) + z = x + y + z; from specification list c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n + 0 = n; (s)

Proof step 9 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, allWritesOKInd(pending, persisted, h0), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: t, pending, persisted

Additional information

Used terms: t, pending, persisted

Proof step 10 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

e = W(t1, l, v), (e.W? ⊃ (e, boolvar) ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending, persisted, (e, boolvar) ' + h0), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ (e.t = t ⊃ (e, boolvar) ';[]).count(λ vpe. ¬ vpe._2) = 1, (e.W? ⊃ (e, boolvar) ';[]).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: allWritesOKInd-recW: ⊦ allWritesOKInd(pending, persisted, (W(t, l, v), b) ' + h) ↔ ¬ pending(t) ∧ (b → persisted(l)) ∧ allWritesOKInd(pending(t := ¬ b), persisted(l := b), h); ⊦ W(t, l, v).t = t; (s) ⊦ a '.filter(p) = (p(a) ⊃ a ';[]); (s) ⊦ W(t, l, v).W?; (s) a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 11 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

(t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending(t1 := ¬ boolvar), persisted(l := boolvar), h0), ¬ pending(t1), boolvar → persisted(l), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) = 0 ∧ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 1, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) = 1

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 12 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

(t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending(t1 := ¬ boolvar), persisted(l := boolvar), h0), ¬ pending(t1), boolvar → persisted(l), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) = 1

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: t1 = t

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ [].count(p) = 0; (s) Used formulas: t1 = t

Proof step 13 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

t1 = t, (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending(t1 := ¬ boolvar), persisted(l := boolvar), h0), ¬ pending(t1), boolvar → persisted(l), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ a '.count(p) = (p(a) ⊃ 1;0); (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≠ 0; (s)

Proof step 14 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending(t := true), persisted(l := false), h0), ¬ pending(t), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: filter-count, name:count-zero-filter-nil, seq: ⊦ x.count(p) = 0 ↔ x.filter(p) = [], subst:[x, p] → [h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t), λ vpe. ¬ vpe._2]; ['a] → [(RWevent, Bool)], <all paths>

Additional information

Used terms: h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t), λ vpe. ¬ vpe._2

Proof step 15 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

⊦ x.count(p) = 0 ↔ x.filter(p) = []

A spec-lemma

Proof step 16 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).filter(λ vpe. ¬ vpe._2) ≠ [], allWritesOKInd(pending(t := true), persisted(l := false), h0), ¬ pending(t), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list, name:some-element, seq: ⊦ x = [] ↔ ¬(∃ a. a ∈ x), subst:[x] → [h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).filter(λ vpe. ¬ vpe._2)]; ['a] → [(RWevent, Bool)], <all paths>

Additional information

Used terms: h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).filter(λ vpe. ¬ vpe._2)

Proof step 17 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

⊦ x = [] ↔ ¬(∃ a. a ∈ x)

A spec-lemma

Proof step 18 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

allWritesOKInd(pending(t := true), persisted(l := false), h0), ¬ pending(t), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0), ∃ a. a ∈ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).filter(λ vpe. ¬ vpe._2) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ x.filter(p) ↔ a ∈ x ∧ p(a); (s)

Proof step 19 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

vpe._1.W?, allWritesOKInd(pending(vpe._1.t := true), persisted(l := false), h0), vpe ∈ h0, ¬ pending(vpe._1.t), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ vpe._2

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

term: vpe._1

Additional information

Used terms: e, boolvar

Proof step 20 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

¬ boolvar, vpe = (e, boolvar), e.W?, allWritesOKInd(pending(e.t := true), persisted(l := false), h0), vpe ∈ h0, ¬ pending(e.t), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 21 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

e.W?, allWritesOKInd(pending(e.t := true), persisted(l := false), h0), (e, false) ∈ h0, ¬ pending(e.t), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: RWevent, name:elim-W, seq: ⊦ e.W? → ((t = e.t ∧ l = e.l) ∧ v = e.v ↔ e = W(t, l, v)), subst:[e, t, l, v] → [e, t, l0, v]

Additional information

Used simplifier rules: ⊦ e.W? → ((t = e.t ∧ l = e.l) ∧ v = e.v ↔ e = W(t, l, v)); (elim) Used terms: e, t, l0, v

Proof step 22 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

e = W(t, l0, v), e.W?, allWritesOKInd(pending(t := true), persisted(l := false), h0), (e, false) ∈ h0, ¬ pending(t), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ W(t, l, v).W?; (s)

Proof step 23 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

allWritesOKInd(pending(t := true), persisted(l := false), h0), (W(t, l0, v), false) ∈ h0, ¬ pending(t), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:allwritesOK-ind-pending-not-in, seq: ⊦ pending(t) ∧ allWritesOKInd(pending, persisted, h) → ¬ (W(t, l, v), false) ∈ h, subst:[t, persisted, l, v, pending, h] → [t, persisted(l := false), l0, v, pending(t := true), h0], <all paths>

Additional information

Used terms: persisted(l := false), v, pending(t := true), l0, t, h0

Proof step 24 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

⊦ pending(t) ∧ allWritesOKInd(pending, persisted, h) → ¬ (W(t, l, v), false) ∈ h

Lemma allwritesOK-ind-pending-not-in

Proof step 25 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

(t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, allWritesOKInd(pending(t1 := ¬ boolvar), persisted(l := boolvar), h0), ¬ pending(t1), boolvar → persisted(l), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) = 0 ∧ (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) = 1

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 26 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

(t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, allWritesOKInd(pending(t1 := ¬ boolvar), persisted(l := boolvar), h0), ¬ pending(t1), boolvar → persisted(l), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: t, pending(t1 := ¬ boolvar), persisted(l := boolvar)

Additional information

Used terms: t, pending(t1 := ¬ boolvar), persisted(l := boolvar)

Proof step 27 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

(t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) ≠ 1, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, allWritesOKInd(pending(t1 := ¬ boolvar), persisted(l := boolvar), h0), ¬ pending(t1), boolvar → persisted(l), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 28 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

t1 = t, (W(t, l, v), boolvar) '.count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) ≠ 1, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, allWritesOKInd(pending(t1 := ¬ boolvar), persisted(l := boolvar), h0), ¬ pending(t1), boolvar → persisted(l), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n + 0 = n; (s) ⊦ a '.count(p) = (p(a) ⊃ 1;0); (s)

Proof step 29 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, allWritesOKInd(pending(t := false), persisted(l := true), h0), persisted(l), ¬ pending(t), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: t, pending(t := false), persisted(l := true)

Additional information

Used terms: t, pending(t := false), persisted(l := true)

Proof step 30 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

[].count(λ vpe. ¬ vpe._2) + h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, (t1 = t ⊃ (W(t, l, v), boolvar) ';[]).count(λ vpe. ¬ vpe._2) ≠ 1, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, t1 ≠ t, allWritesOKInd(pending(t1 := ¬ boolvar), persisted(l := boolvar), h0), ¬ pending(t1), boolvar → persisted(l), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ [].count(p) = 0; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n + 0 = n; (s)

Proof step 31 from proof allwritesOK-ind-at-most-one-write (previous step) (back to tree)

h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 1, h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t).count(λ vpe. ¬ vpe._2) ≠ 0, t1 ≠ t, allWritesOKInd(pending(t1 := ¬ boolvar), persisted(l := boolvar), h0), ¬ pending(t1), boolvar → persisted(l), ∀ t0, pending0, persisted0. h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 0 ∨ h0.filter(λ vpe. vpe._1.W?).filter(λ vpe. vpe._1.t = t0).count(λ vpe. ¬ vpe._2) = 1 ∨ ¬ allWritesOKInd(pending0, persisted0, h0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: t, pending(t1 := ¬ boolvar), persisted(l := boolvar)

Additional information

Used terms: t, pending(t1 := ¬ boolvar), persisted(l := boolvar)

Proof Tree for allwritesOK-ind-at-most-one-write (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree