[Imprint][Data Privacy Statement][E-Mail]



View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 1 from proof del-eq (conclusion) (back to tree)

⊦ a ∈ s ∧ a ∈ s0 → (s -- a = s0 -- a ↔ s = s0)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof del-eq (previous step) (back to tree)

¬(s -- a = s0 -- a ↔ s = s0), a ∈ s, a ∈ s0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; ⊦ ¬ a ∈ ∅; (s) ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s) ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic

Proof step 3 from proof del-eq (previous step) (back to tree)

a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0 ⊦ (∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0)) ↔ (∀ a. a ∈ s ↔ a ∈ s0)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 4 from proof del-eq (previous step) (back to tree)

a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ ∀ a. a ∈ s ↔ a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 5 from proof del-eq (previous step) (back to tree)

¬(a2 ∈ s ↔ a2 ∈ s0), a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: a2 ∈ s

Additional information

Used formulas: a2 ∈ s

Proof step 6 from proof del-eq (previous step) (back to tree)

¬(a2 ∈ s ↔ a2 ∈ s0), a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, a2 ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 7 from proof del-eq (previous step) (back to tree)

a2 ∈ s, a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0, ¬ a2 ∈ s0, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: a2

Additional information

Used terms: a2

Proof step 8 from proof del-eq (previous step) (back to tree)

¬(a2 ∈ s ↔ a2 ∈ s0), a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, ¬ a2 ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 9 from proof del-eq (previous step) (back to tree)

a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0, a2 ∈ s0, ¬ a2 ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: a2

Additional information

Used terms: a2

Proof step 10 from proof del-eq (previous step) (back to tree)

a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦ ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 11 from proof del-eq (previous step) (back to tree)

¬(a2 ∈ s ↔ a2 ∈ s0), a2 ≠ a, a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0, ∀ a. a ∈ s ↔ a ∈ s0 ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: a2 = a

Additional information

Used formulas: a2 = a

Proof step 12 from proof del-eq (previous step) (back to tree)

¬(a2 ∈ s ↔ a2 ∈ s0), a2 ≠ a, a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 13 from proof del-eq (previous step) (back to tree)

¬(a2 ∈ s ↔ a2 ∈ s0), a2 ≠ a, a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0, ∀ a. a ∈ s ↔ a ∈ s0 ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: a2 ∈ s

Additional information

Used formulas: a2 ∈ s

Proof step 14 from proof del-eq (previous step) (back to tree)

¬(a2 ∈ s ↔ a2 ∈ s0), a2 ≠ a, a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, a2 ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 15 from proof del-eq (previous step) (back to tree)

a2 ≠ a, a2 ∈ s, a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0, ¬ a2 ∈ s0, ∀ a. a ∈ s ↔ a ∈ s0 ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: a2

Additional information

Used terms: a2

Proof step 16 from proof del-eq (previous step) (back to tree)

¬(a2 ∈ s ↔ a2 ∈ s0), a2 ≠ a, a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, ¬ a2 ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 17 from proof del-eq (previous step) (back to tree)

a2 ≠ a, a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0, a2 ∈ s0, ¬ a2 ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: a2

Additional information

Used terms: a2

Proof Tree for del-eq (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