Proof step 1 from proof oplus-v-append (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  v ⊕ (fowl0 + fowl1) ≠ (v ⊕ fowl0) ⊕ fowl1 ⊦

Rule argument

Additional information

Proof step 2 from proof oplus-v-append (previous step) (back to tree)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

• Proof step 3:

  v ⊕ ((fow + fowl) + fowl1) ≠ (v ⊕ (fow + fowl)) ⊕ fowl1, ∀ v, fowl1. v ⊕ (fowl + fowl1) = (v ⊕ fowl) ⊕ fowl1 ⊦

Rule argument

Additional information

Used simplifier rules: overwrite-base: ⊦ v ⊕ [] = v; oplus-same-writes-eq: ⊦ fowl0 |W = fowl1 |W → (v ⊕ fowl0 = v ⊕ fowl1 ↔ true); List('a) freely generated by [], +;; ⊦ [] + x = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 3 from proof oplus-v-append (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  v ⊕ ((fow ' + fowl) + fowl1) ≠ (v ⊕ (fow ' + fowl)) ⊕ fowl1, ∀ v, fowl1. v ⊕ (fowl + fowl1) = (v ⊕ fowl) ⊕ fowl1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 4 from proof oplus-v-append (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 5:

  fow = FO(t), v ⊕ ((fow ' + fowl) + fowl1) ≠ (v ⊕ (fow ' + fowl)) ⊕ fowl1, ∀ v, fowl1. v ⊕ (fowl + fowl1) = (v ⊕ fowl) ⊕ fowl1 ⊦

• Proof step 7:

  fow = W(v0, t), v ⊕ ((fow ' + fowl) + fowl1) ≠ (v ⊕ (fow ' + fowl)) ⊕ fowl1, ∀ v, fowl1. v ⊕ (fowl + fowl1) = (v ⊕ fowl) ⊕ fowl1 ⊦

Rule argument

Additional information

Used simplifier rules: FOW freely generated by FO, W;;

Proof step 5 from proof oplus-v-append (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  v ⊕ (fowl + fowl1) ≠ (v ⊕ fowl) ⊕ fowl1, ∀ v, fowl1. v ⊕ (fowl + fowl1) = (v ⊕ fowl) ⊕ fowl1 ⊦

Rule argument

Additional information

Used simplifier rules: overwrite-rec1: ⊦ v ⊕ (FO(t) ' + fowl) = v ⊕ fowl; a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 6 from proof oplus-v-append (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: v, fowl1

Proof step 7 from proof oplus-v-append (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  v0 ⊕ (fowl + fowl1) ≠ (v0 ⊕ fowl) ⊕ fowl1, ∀ v, fowl1. v ⊕ (fowl + fowl1) = (v ⊕ fowl) ⊕ fowl1 ⊦

Rule argument

Additional information

Used simplifier rules: overwrite-rec2: ⊦ v ⊕ (W(v0, t) ' + fowl) = v0 ⊕ fowl; a: ⊦ (x + y) + z = x + y + z; from specification list

Proof step 8 from proof oplus-v-append (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: v0, fowl1