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Proof step 1 from proof Sep-forall-prove-imp (conclusion) (back to tree)

∀ n0. n0 < m → nP0(n + n0) ⊆ nP1(n + n0) ⊦ nP0.⊗_(n, m) ⊆ nP1.⊗_(n, m)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof Sep-forall-prove-imp (previous step) (back to tree)

¬ nP0.⊗_(n, m) ⊆ nP1.⊗_(n, m), ∀ n0. n0 < m → nP0(n + n0) ⊆ nP1(n + n0) ⊦

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Rule argument

term: m

Additional information

Used simplifier rules: Sep-base: ⊦ nP.⊗_(n, 0) = emp; Nat freely generated by 0, +1; ⊦ hP ⊆ hP; (s)

Proof step 3 from proof Sep-forall-prove-imp (previous step) (back to tree)

¬ nP0.⊗_(n, m0.+1) ⊆ nP1.⊗_(n, m0.+1), ∀ n0. n0 < m0.+1 → nP0(n + n0) ⊆ nP1(n + n0), ∀ nP0, n, nP1. ¬(¬ nP0.⊗_(n, m0) ⊆ nP1.⊗_(n, m0) ∧ (∀ n0. n0 < m0 → nP0(n + n0) ⊆ nP1(n + n0))) ⊦

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 ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m.+1 = m + 1; (s)

Proof step 4 from proof Sep-forall-prove-imp (previous step) (back to tree)

¬ nP0.⊗_(n, m0 + 1) ⊆ nP1.⊗_(n, m0 + 1), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0), ∀ nP, n0, nP2. nP.⊗_(n0, m0) ⊆ nP2.⊗_(n0, m0) ∨ ¬(∀ n1. n1 < m0 → nP(n0 + n1) ⊆ nP2(n0 + n1)) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name: Sep-rec, seq: ⊦ nP.⊗_(n, m + 1) = nP(n) * nP.⊗_(n + 1, m), subst:[m, n, nP] → [m0, n, nP0], <all paths>

Additional information

Used terms: m0, n, nP0

Proof step 5 from proof Sep-forall-prove-imp (previous step) (back to tree)

⊦ nP.⊗_(n, m + 1) = nP(n) * nP.⊗_(n + 1, m)

Lemma Sep-rec

Proof step 6 from proof Sep-forall-prove-imp (previous step) (back to tree)

¬ nP0(n) * nP0.⊗_(n + 1, m0) ⊆ nP1.⊗_(n, m0 + 1), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0), ∀ nP, n0, nP2. nP.⊗_(n0, m0) ⊆ nP2.⊗_(n0, m0) ∨ ¬(∀ n1. n1 < m0 → nP(n0 + n1) ⊆ nP2(n0 + n1)) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name: Sep-rec, seq: ⊦ nP.⊗_(n, m + 1) = nP(n) * nP.⊗_(n + 1, m), subst:[m, n, nP] → [m0, n, nP1], <all paths>

Additional information

Used terms: m0, n, nP1

Proof step 7 from proof Sep-forall-prove-imp (previous step) (back to tree)

⊦ nP.⊗_(n, m + 1) = nP(n) * nP.⊗_(n + 1, m)

Lemma Sep-rec

Proof step 8 from proof Sep-forall-prove-imp (previous step) (back to tree)

¬ nP0(n) * nP0.⊗_(n + 1, m0) ⊆ nP1(n) * nP1.⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0), ∀ nP, n0, nP2. nP.⊗_(n0, m0) ⊆ nP2.⊗_(n0, m0) ∨ ¬(∀ n1. n1 < m0 → nP(n0 + n1) ⊆ nP2(n0 + n1)) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: heap-sep, name: sep-inj2-imp, seq: ⊦ hP0 ⊆ hP1 ∧ hP2 ⊆ hP3 → hP0 * hP2 ⊆ hP1 * hP3, subst:[hP3, hP1, hP2, hP0] → [nP1.⊗_(n + 1, m0), nP1(n), nP0.⊗_(n + 1, m0), nP0(n)], <all paths>

Additional information

Used terms: nP1.⊗_(n + 1, m0), nP1(n), nP0.⊗_(n + 1, m0), nP0(n)

Proof step 9 from proof Sep-forall-prove-imp (previous step) (back to tree)

⊦ hP0 ⊆ hP1 ∧ hP2 ⊆ hP3 → hP0 * hP2 ⊆ hP1 * hP3

A spec-lemma

Proof step 10 from proof Sep-forall-prove-imp (previous step) (back to tree)

¬ nP0(n) * nP0.⊗_(n + 1, m0) ⊆ nP1(n) * nP1.⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0), ∀ nP, n0, nP2. nP.⊗_(n0, m0) ⊆ nP2.⊗_(n0, m0) ∨ ¬(∀ n1. n1 < m0 → nP(n0 + n1) ⊆ nP2(n0 + n1)) ⊦ nP0(n) ⊆ nP1(n) ∧ nP0.⊗_(n + 1, m0) ⊆ nP1.⊗_(n + 1, m0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: nP0(n) ⊆ nP1(n)

Additional information

Used formulas: nP0(n) ⊆ nP1(n)

Proof step 11 from proof Sep-forall-prove-imp (previous step) (back to tree)

nP0(n) ⊆ nP1(n), ¬ nP0(n) * nP0.⊗_(n + 1, m0) ⊆ nP1(n) * nP1.⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0), ∀ nP, n0, nP2. nP.⊗_(n0, m0) ⊆ nP2.⊗_(n0, m0) ∨ ¬(∀ n1. n1 < m0 → nP(n0 + n1) ⊆ nP2(n0 + n1)) ⊦ nP0(n) ⊆ nP1(n) ∧ nP0.⊗_(n + 1, m0) ⊆ nP1.⊗_(n + 1, m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 12 from proof Sep-forall-prove-imp (previous step) (back to tree)

nP0(n) ⊆ nP1(n), ¬ nP0(n) * nP0.⊗_(n + 1, m0) ⊆ nP1(n) * nP1.⊗_(n + 1, m0), ¬ nP0.⊗_(n + 1, m0) ⊆ nP1.⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0), ∀ nP, n0, nP2. nP.⊗_(n0, m0) ⊆ nP2.⊗_(n0, m0) ∨ ¬(∀ n1. n1 < m0 → nP(n0 + n1) ⊆ nP2(n0 + n1)) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 5 terms: nP0, n + 1, nP1 abort: F discard: T computed: T

Additional information

Used terms: nP0, n + 1, nP1

Proof step 13 from proof Sep-forall-prove-imp (previous step) (back to tree)

nP0(n) ⊆ nP1(n), ¬ nP0(n) * nP0.⊗_(n + 1, m0) ⊆ nP1(n) * nP1.⊗_(n + 1, m0), ¬ nP0.⊗_(n + 1, m0) ⊆ nP1.⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0), nP0.⊗_(n + 1, m0) ⊆ nP1.⊗_(n + 1, m0) ∨ ¬(∀ n1. n1 < m0 → nP0((n + 1) + n1) ⊆ nP1((n + 1) + n1)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 14 from proof Sep-forall-prove-imp (previous step) (back to tree)

n1 < m0, nP0(n) ⊆ nP1(n), ¬ nP0((n + 1) + n1) ⊆ nP1((n + 1) + n1), ¬ nP0.⊗_(n + 1, m0) ⊆ nP1.⊗_(n + 1, m0), ¬ nP0(n) * nP0.⊗_(n + 1, m0) ⊆ nP1(n) * nP1.⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n1 + 1

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1 + 1

Proof step 15 from proof Sep-forall-prove-imp (previous step) (back to tree)

¬ nP0(n) ⊆ nP1(n), ¬ nP0(n) * nP0.⊗_(n + 1, m0) ⊆ nP1(n) * nP1.⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0), ∀ nP, n0, nP2. nP.⊗_(n0, m0) ⊆ nP2.⊗_(n0, m0) ∨ ¬(∀ n1. n1 < m0 → nP(n0 + n1) ⊆ nP2(n0 + n1)) ⊦ nP0(n) ⊆ nP1(n) ∧ nP0.⊗_(n + 1, m0) ⊆ nP1.⊗_(n + 1, m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 16 from proof Sep-forall-prove-imp (previous step) (back to tree)

¬ nP0(n) ⊆ nP1(n), ¬ nP0(n) * nP0.⊗_(n + 1, m0) ⊆ nP1(n) * nP1.⊗_(n + 1, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) ⊆ nP1(n + n0), ∀ nP, n0, nP2. nP.⊗_(n0, m0) ⊆ nP2.⊗_(n0, m0) ∨ ¬(∀ n1. n1 < m0 → nP(n0 + n1) ⊆ nP2(n0 + n1)) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: 0

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s) ⊦ n + 0 = n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: 0

Proof Tree for Sep-forall-prove-imp (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