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Proof step 1 from proof addvec-assoc (conclusion) (back to tree)

⊦ # vec1 = # vec2 ∧ # vec1 = # vec2 → (vec1 + vec2) + vec3 = vec1 + vec2 + vec3

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof addvec-assoc (previous step) (back to tree)

# vec1 = # vec2, (vec1 + vec2) + vec3 ≠ vec1 + vec2 + vec3 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: array, name:Extension, seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]), subst:[ar1, ar2] → [(vec1 + vec2) + vec3, vec1 + vec2 + vec3]; ['a] → [ring], <all paths>

Additional information

Used terms: (vec1 + vec2) + vec3, vec1 + vec2 + vec3

Proof step 3 from proof addvec-assoc (previous step) (back to tree)

⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])

A spec-lemma

Proof step 4 from proof addvec-assoc (previous step) (back to tree)

# vec1 = # vec2, #((vec1 + vec2) + vec3) ≠ #(vec1 + vec2 + vec3) ∨ (∃ n. ¬(n < #((vec1 + vec2) + vec3) → ((vec1 + vec2) + vec3)[n] = (vec1 + vec2 + vec3)[n])) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: addvec-size: ⊦ #(vec1 + vec2) = # vec1; addmvec-get: ⊦ n < # vec1 → (vec1 + vec2)[n] = vec1[n] + vec2[n]; addassoc: ⊦ (a + b) + c = a + b + c; from specification ring ⊦ m < n → n ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 5 from proof addvec-assoc (previous step) (back to tree)

# vec1 = # vec2, ((vec1 + vec2) + vec3)[n] ≠ vec1[n] + vec2[n] + vec3[n], n < # vec2 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:addmvec-get, seq: ⊦ n < # vec1 → (vec1 + vec2)[n] = vec1[n] + vec2[n], subst:[vec1, vec2, n] → [vec1 + vec2, vec3, n], <all paths>

Additional information

Used simplifier rules: addvec-size: ⊦ #(vec1 + vec2) = # vec1; addvec-size: ⊦ #(vec1 + vec2) = # vec1; addmvec-get: ⊦ n < # vec1 → (vec1 + vec2)[n] = vec1[n] + vec2[n]; addassoc: ⊦ (a + b) + c = a + b + c; from specification ring addcom: ⊦ a + b = b + a; from specification ring ⊦ m < n ↔ ¬ n ≤ m; (ss) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ ¬ n < n; (s) Used terms: vec1 + vec2, vec3, n

Proof step 6 from proof addvec-assoc (previous step) (back to tree)

⊦ n < # vec1 → (vec1 + vec2)[n] = vec1[n] + vec2[n]

Lemma addmvec-get

Proof Tree for addvec-assoc (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