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Proof step 1 from proof Sep-split (conclusion) (back to tree)

⊦ m0 ≤ m → nP ⊗_(n, m) = nP ⊗_(n, m0) * nP ⊗_(n + m0, m - m0)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof Sep-split (previous step) (back to tree)

nP ⊗_(n, m) ≠ nP ⊗_(n, m0) * nP ⊗_(n + m0, m - m0), m0 ≤ m ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [m, m0, n0]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m, m0, n0

Proof step 3 from proof Sep-split (previous step) (back to tree)

m = n0 + m0, nP ⊗_(n, m) ≠ nP ⊗_(n, m0) * nP ⊗_(n + m0, n0), m0 ≤ m ⊦

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 ≤ m + n; (s)

Proof step 4 from proof Sep-split (previous step) (back to tree)

nP ⊗_(n, n0 + m0) ≠ nP ⊗_(n, m0) * nP ⊗_(n + m0, n0) ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: m0

Additional information

Used simplifier rules: Sep-base: ⊦ nP ⊗_(n, 0) = emp; nat freely generated by 0, +1;; ⊦ n + 0 = n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ emp * hP = hP; (s) sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep

Proof step 5 from proof Sep-split (previous step) (back to tree)

nP ⊗_(n, n0 + m +1) ≠ nP ⊗_(n, m +1) * nP ⊗_(n + m +1, n0), ∀ nP, n, n0. nP ⊗_(n, n0 + m) = nP ⊗_(n, m) * nP ⊗_(n + m, n0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m +1 = m + 1; (s)

Proof step 6 from proof Sep-split (previous step) (back to tree)

nP ⊗_(n, n0 + m + 1) ≠ nP ⊗_(n, m + 1) * nP ⊗_(n + m + 1, n0), ∀ nP, n, n0. nP ⊗_(n, n0 + m) = nP ⊗_(n, m) * nP ⊗_(n + m, n0) ⊦

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] → [m + n0, n, nP], <all paths>

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: m + n0, n, nP

Proof step 7 from proof Sep-split (previous step) (back to tree)

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

Lemma Sep-rec

Proof step 8 from proof Sep-split (previous step) (back to tree)

nP(n) * nP ⊗_(n + 1, m + n0) ≠ nP ⊗_(n, m + 1) * nP ⊗_(n + m + 1, n0), ∀ nP, n, n0. nP ⊗_(n, n0 + m) = nP ⊗_(n, m) * nP ⊗_(n + m, n0) ⊦

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] → [m, n, nP], <all paths>

Additional information

Used terms: m, n, nP

Proof step 9 from proof Sep-split (previous step) (back to tree)

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

Lemma Sep-rec

Proof step 10 from proof Sep-split (previous step) (back to tree)

nP(n) * nP ⊗_(n + 1, m + n0) ≠ (nP(n) * nP ⊗_(n + 1, m)) * nP ⊗_(n + m + 1, n0), ∀ nP, n, n0. nP ⊗_(n, n0 + m) = nP ⊗_(n, m) * nP ⊗_(n + m, n0) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 2 terms: nP, n + 1, n0 abort: F discard: T computed: F

Additional information

Used terms: nP, n + 1, n0

Proof step 11 from proof Sep-split (previous step) (back to tree)

nP ⊗_(n + 1, n0 + m) = nP ⊗_(n + 1, m) * nP ⊗_((n + 1) + m, n0), nP(n) * nP ⊗_(n + 1, m + n0) ≠ (nP(n) * nP ⊗_(n + 1, m)) * nP ⊗_(n + m + 1, n0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-basic, name:c, seq: ⊦ m + n = n + m, subst:[n, m] → [m, n0], <all paths>

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic Used terms: m, n0

Proof step 12 from proof Sep-split (previous step) (back to tree)

⊦ m + n = n + m

A spec-lemma

Proof step 13 from proof Sep-split (previous step) (back to tree)

nP ⊗_(n + 1, m + n0) = nP ⊗_(n + 1, m) * nP ⊗_((n + 1) + m, n0), nP(n) * nP ⊗_(n + 1, m + n0) ≠ (nP(n) * nP ⊗_(n + 1, m)) * nP ⊗_(n + m + 1, n0) ⊦

User applied rule subst equation

And closed the goal!

Rule argument

pos: left 1, flags: discard

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep

Proof Tree for Sep-split (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