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Proof step 1 from proof addmx-com (conclusion) (back to tree)

⊦ #r mx1 = #r mx2 ∧ #c mx1 = #c mx2 → mx1 + mx2 = mx2 + mx1

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof addmx-com (previous step) (back to tree)

#c mx1 = #c mx2, #r mx1 = #r mx2, mx1 + mx2 ≠ mx2 + mx1 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: matrix, name:extension, seq:⊦ mx1 = mx2 ↔ #r mx1 = #r mx2 ∧ #c mx1 = #c mx2 ∧ (∀ r, c. r < #r mx1 ∧ c < #c mx1 → mx1[r, c] = mx2[r, c]), subst:[mx1, mx2] → [mx1 + mx2, mx2 + mx1]; ['a] → [ring], <all paths>

Additional information

Used simplifier rules: addmx-rowno: ⊦ #r(mx1 + mx2) = #r mx1; addmx-colno: ⊦ #c(mx1 + mx2) = #c mx1; addmx-get: ⊦ r < #r mx1 ∧ c < #c mx1 → (mx1 + mx2)[r, c] = mx1[r, c] + mx2[r, c]; f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s) addcom: ⊦ a + b = b + a; from specification ring Used terms: mx1 + mx2, mx2 + mx1

Proof step 3 from proof addmx-com (previous step) (back to tree)

⊦ mx1 = mx2 ↔ #r mx1 = #r mx2 ∧ #c mx1 = #c mx2 ∧ (∀ r, c. r < #r mx1 ∧ c < #c mx1 → mx1[r, c] = mx2[r, c])

A spec-lemma

Proof Tree for addmx-com (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