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Proof step 1 from proof invmult-r (conclusion) (back to tree)

⊦ b * O- a = O-(b * a)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof invmult-r (previous step) (back to tree)

b * O- a ≠ O-(b * a) ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: b * (O- a + a) = O

Additional information

Used simplifier rules: addinv-l: ⊦ O- a + a = O; addassoc: ⊦ (a + b) + c = a + b + c; addcom: ⊦ a + b = b + a; multzero-r: ⊦ a * O = O; multassoc: ⊦ (a * b) * c = a * b * c; Used formulas: b * (O- a + a) = O

Proof step 3 from proof invmult-r (previous step) (back to tree)

b * (O- a + a) = O, b * O- a ≠ O-(b * a) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:distr-r, seq: ⊦ a * (b + c) = a * b + a * c, subst:[a, b, c] → [b, O- a, a], <all paths>

Additional information

Used terms: b, O- a, a

Proof step 4 from proof invmult-r (previous step) (back to tree)

⊦ a * (b + c) = a * b + a * c

Lemma distr-r

Proof step 5 from proof invmult-r (previous step) (back to tree)

b * O- a + b * a = O, b * O- a ≠ O-(b * a) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:addinv-unique, seq: ⊦ a + b = O ↔ b = O- a, subst:[a, b] → [b * O- a, b * a], <all paths>

Additional information

Used terms: b * O- a, b * a

Proof step 6 from proof invmult-r (previous step) (back to tree)

⊦ a + b = O ↔ b = O- a

Lemma addinv-unique

Proof step 7 from proof invmult-r (previous step) (back to tree)

b * a = O-(b * O- a), b * O- a ≠ O-(b * a) ⊦

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 1, flags: discard

Additional information

Proof step 8 from proof invmult-r (previous step) (back to tree)

b * O- a ≠ O- O-(b * O- a) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:invtwice, seq: ⊦ O- O- a = a, subst:[a] → [b * O- a], <all paths>

Additional information

Used terms: b * O- a

Proof step 9 from proof invmult-r (previous step) (back to tree)

⊦ O- O- a = a

Lemma invtwice

Proof Tree for invmult-r (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