Proof step 1 from proof idexcept-n-copy-other (conclusion) (back to tree)

⊦ m2 ≤ m1 ∧ m1 + n ≤ m2 + m → (id○(ar0, copy(ar, m0, ar1, m1, n), m2, m) ↔ id○(ar0, ar1, m2, m))

System applied rule normalize

And obtained the following new goals

Proof step 2:
¬(id○(ar0, copy(ar, m0, ar1, m1, n), m2, m) ↔ id○(ar0, ar1, m2, m)), m2 ≤ m1, m1 + n ≤ m2 + m ⊦

Rule argument

Additional information

Proof step 2 from proof idexcept-n-copy-other (previous step) (back to tree)

¬(id○(ar0, copy(ar, m0, ar1, m1, n), m2, m) ↔ id○(ar0, ar1, m2, m)), m2 ≤ m1, m1 + n ≤ m2 + m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# ar0 = # ar1, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m ⊦ (∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1))) ↔ (∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0))

Rule argument

Additional information

Used simplifier rules: size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0; Idexcept-n: ⊦ id○(ar0, ar1, m, n) ↔ # ar0 = # ar1 ∧ m + n ≤ # ar0 ∧ (∀ m0. (m0 < m ∨ m + n ≤ m0) ∧ m0 < # ar0 → ar0(m0) = ar1(m0)); at-copy: ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)(m0) = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1(m0);ar0(m0 + n - n0)); let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat

Proof step 3 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m ⊦ (∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1))) ↔ (∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0))

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 4:
# ar0 = # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)) ⊦ ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0)
Proof step 10:
# ar0 = # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0) ⊦ ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1))

Rule argument

pos: right 1

Additional information

Proof step 4 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)) ⊦ ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 5 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < # ar1, m3 < m2, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < m2 ∨ m2 + m ≤ m3 ⊦
Proof step 8:
# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < # ar1, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m, ¬ m3 < m2, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Rule argument

fma: m3 < m2

Additional information

Used formulas: m3 < m2

Proof step 6 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < # ar1, m3 < m2, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < m2, m3 < # ar1, m3 < m1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat

Proof step 7 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < m2, m3 < # ar1, m3 < m1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: m3

Additional information

Used terms: m3

Proof step 8 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < # ar1, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m, ¬ m3 < m2, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m1 + n < # ar1, m3 < # ar1, m2 < # ar1, m2 + m < # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 ≤ m3, m2 + m ≤ m3, m1 + n ≤ m3, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 9 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m1 + n < # ar1, m3 < # ar1, m2 < # ar1, m2 + m < # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 ≤ m3, m2 + m ≤ m3, m1 + n ≤ m3, ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 12 terms: m3

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: m3

Proof step 10 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0) ⊦ ∀ m3. (m3 < m2 ∨ m2 + m ≤ m3) ∧ m3 < # ar1 → ar0(m3) = (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# ar0 = # ar1, ar0(m3) ≠ (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 11 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 12:
# ar0 = # ar1, ar0(m3) ≠ (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < # ar1, m3 < m2, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0), m3 < m2 ∨ m2 + m ≤ m3 ⊦
Proof step 14:
# ar0 = # ar1, ar0(m3) ≠ (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < # ar1, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m, ¬ m3 < m2, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Rule argument

fma: m3 < m2

Additional information

Used formulas: m3 < m2

Proof step 12 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < # ar1, m3 < m2, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < m2, m3 < # ar1, m3 < m1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat

Proof step 13 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m3 < m2, m3 < # ar1, m3 < m1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 + m ≤ # ar1, m1 + n ≤ # ar1, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: m3

Additional information

Used terms: m3

Proof step 14 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ (m3 < m1 ∨ m1 + n ≤ m3 ⊃ ar1(m3);ar(m3 + m0 - m1)), m3 < # ar1, m1 + n ≤ # ar1, m2 + m ≤ # ar1, m2 ≤ m1, m1 + n ≤ m2 + m, ¬ m3 < m2, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0), m3 < m2 ∨ m2 + m ≤ m3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m1 + n < # ar1, m3 < # ar1, m2 < # ar1, m2 + m < # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 ≤ m3, m2 + m ≤ m3, m1 + n ≤ m3, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0) ⊦

Rule argument

Additional information

Proof step 15 from proof idexcept-n-copy-other (previous step) (back to tree)

# ar0 = # ar1, ar0(m3) ≠ ar1(m3), m1 + n < # ar1, m3 < # ar1, m2 < # ar1, m2 + m < # ar1, m1 + n ≤ m2 + m, m2 ≤ m1, m2 ≤ m3, m2 + m ≤ m3, m1 + n ≤ m3, ∀ m0. (m0 < m2 ∨ m2 + m ≤ m0) ∧ m0 < # ar1 → ar0(m0) = ar1(m0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 12 terms: m3

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: m3

Proof Tree for idexcept-n-copy-other (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.