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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 unall (conclusion) (back to tree)

⊦ s = s0 ∪ s ↔ (∀ a. a ∈ s0 → a ∈ s)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: unrax: ⊦ a ∈ s ∪ s0 ↔ a ∈ s ∨ a ∈ s0; una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; ⊦ ¬ a ∈ ∅; (s) ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s) ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification polyinfset-basic

Proof step 2 from proof unall (previous step) (back to tree)

a ∈ s0, a0 ∈ s0, ¬ a ∈ s, ∀ a. a ∈ s0 → a ∈ s ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a

Additional information

Used terms: a

Proof Tree for unall (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