Proof step 1 from proof map-in (conclusion) (back to tree)

⊦ b ∈ x map f ↔ (∃ a. b = f(a) ∧ a ∈ x)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b ∈ (a + x0) map f ↔ (∃ a0. b = f(a0) ∧ a0 ∈ a + x0)

Rule argument

term: x

Additional information

Used simplifier rules: map-base: ⊦ [] map f = []; in-empty: ⊦ ¬ a ∈ []; list('a) freely generated by [], +;;

Proof step 2 from proof map-in (previous step) (back to tree)

∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b ∈ (a + x0) map f ↔ (∃ a0. b = f(a0) ∧ a0 ∈ a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b = f(a) ∨ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ (a0 = a ∨ a0 ∈ x0))

Rule argument

Additional information

Used simplifier rules: map-rec: ⊦ (a ' + x) map f = f(a) + (x map f); c: ⊦ a + x = a ' + x; in-rec: ⊦ a ∈ a0 ' + x ↔ a = a0 ∨ a ∈ x; a: ⊦ (x + y) + z = x + y + z;

Proof step 3 from proof map-in (previous step) (back to tree)

∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b = f(a) ∨ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ (a0 = a ∨ a0 ∈ x0))

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
b = f(a), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b = f(a) ∨ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ (a0 = a ∨ a0 ∈ x0))
Proof step 6:
b ≠ f(a), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b = f(a) ∨ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ (a0 = a ∨ a0 ∈ x0))

Rule argument

fma: b = f(a)

Additional information

Used formulas: b = f(a)

Proof step 4 from proof map-in (previous step) (back to tree)

b = f(a), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b = f(a) ∨ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ (a0 = a ∨ a0 ∈ x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a0. f(a) = f(a0) ∧ (a0 = a ∨ a0 ∈ x0)

Rule argument

Additional information

Proof step 5 from proof map-in (previous step) (back to tree)

∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a0. f(a) = f(a0) ∧ (a0 = a ∨ a0 ∈ x0)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: a

Additional information

Used terms: a

Proof step 6 from proof map-in (previous step) (back to tree)

b ≠ f(a), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b = f(a) ∨ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ (a0 = a ∨ a0 ∈ x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
b ≠ f(a), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ a0 ∈ x0)

Rule argument

Additional information

Proof step 7 from proof map-in (previous step) (back to tree)

b ≠ f(a), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ a0 ∈ x0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 8:
b ≠ f(a), b ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ a0 ∈ x0)
Proof step 12:
b ≠ f(a), ¬ b ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ a0 ∈ x0)

Rule argument

fma: b ∈ x0 map f

Additional information

Used formulas: b ∈ x0 map f

Proof step 8 from proof map-in (previous step) (back to tree)

b ≠ f(a), b ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ a0 ∈ x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
b ≠ f(a), b ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a0. b = f(a0) ∧ a0 ∈ x0

Rule argument

Additional information

Proof step 9 from proof map-in (previous step) (back to tree)

b ≠ f(a), b ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a0. b = f(a0) ∧ a0 ∈ x0

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 10:
b ≠ f(a), b ∈ x0 map f, b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a0. b = f(a0) ∧ a0 ∈ x0

Rule argument

pos: left 3 terms: b, f

Additional information

Used terms: b, f

Proof step 10 from proof map-in (previous step) (back to tree)

b ≠ f(a), b ∈ x0 map f, b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a0. b = f(a0) ∧ a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
f(a0) ≠ f(a), a0 ∈ x0, f(a0) ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a. f(a0) = f(a) ∧ a ∈ x0

Rule argument

Additional information

Proof step 11 from proof map-in (previous step) (back to tree)

f(a0) ≠ f(a), a0 ∈ x0, f(a0) ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a. f(a0) = f(a) ∧ a ∈ x0

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: a0

Additional information

Used terms: a0

Proof step 12 from proof map-in (previous step) (back to tree)

b ≠ f(a), ¬ b ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ b ∈ x0 map f ↔ (∃ a0. b = f(a0) ∧ a0 ∈ x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
f(a0) ≠ f(a), a0 ∈ x0, ¬ f(a0) ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦

Rule argument

Additional information

Proof step 13 from proof map-in (previous step) (back to tree)

f(a0) ≠ f(a), a0 ∈ x0, ¬ f(a0) ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦

User applied rule all left

And obtained the following new goals

Proof step 14:
f(a0) ≠ f(a), a0 ∈ x0, ¬ f(a0) ∈ x0 map f, f(a0) ∈ x0 map f ↔ (∃ a. f(a0) = f(a) ∧ a ∈ x0), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦

Rule argument

pos: left 4 terms: f(a0), f abort: F discard: F computed: T

Additional information

Used terms: f(a0), f

Proof step 14 from proof map-in (previous step) (back to tree)

f(a0) ≠ f(a), a0 ∈ x0, ¬ f(a0) ∈ x0 map f, f(a0) ∈ x0 map f ↔ (∃ a. f(a0) = f(a) ∧ a ∈ x0), ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
f(a0) ≠ f(a), a0 ∈ x0, ¬ f(a0) ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a. f(a0) = f(a) ∧ a ∈ x0

Rule argument

Additional information

Proof step 15 from proof map-in (previous step) (back to tree)

f(a0) ≠ f(a), a0 ∈ x0, ¬ f(a0) ∈ x0 map f, ∀ b, f. b ∈ x0 map f ↔ (∃ a. b = f(a) ∧ a ∈ x0) ⊦ ∃ a. f(a0) = f(a) ∧ a ∈ x0

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: a0

Additional information

Used terms: a0

Proof Tree for map-in (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.