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

¬ a ∈ s ⊦ dom st1 \ s = dom st2 \ s ∧ a ∈ st1 → a ∈ st2

System applied rule normalize

And obtained the following new goals

Proof step 2:
dom st1 \ s = dom st2 \ s, a ∈ st1, ¬ a ∈ st2, ¬ a ∈ s ⊦

Rule argument

Additional information

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

dom st1 \ s = dom st2 \ s, a ∈ st1, ¬ a ∈ st2, ¬ a ∈ s ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 4:
a ∈ st1, ¬ a ∈ st2, ¬ a ∈ s, ∀ a. a ∈ dom st1 \ s ↔ a ∈ dom st2 \ s ⊦

Rule argument

spec: set-basic, name:Extension, seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2), subst:[s1, s2] → [dom st1 \ s, dom st2 \ s]; [A] → [A], lhs:dom st1 \ s = dom st2 \ s, <all paths>

Additional information

Used terms: dom st1 \ s, dom st2 \ s

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

⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)

A spec-lemma

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

a ∈ st1, ¬ a ∈ st2, ¬ a ∈ s, ∀ a. a ∈ dom st1 \ s ↔ a ∈ dom st2 \ s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ st1, ¬ a ∈ st2, ¬ a ∈ s, ∀ a0. ¬ a0 ∈ s → (a0 ∈ st1 ↔ a0 ∈ st2) ⊦

Rule argument

Additional information

Used simplifier rules: Dom: ⊦ a ∈ dom st ↔ a ∈ st; ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; (s)

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

a ∈ st1, ¬ a ∈ st2, ¬ a ∈ s, ∀ a0. ¬ a0 ∈ s → (a0 ∈ st1 ↔ a0 ∈ st2) ⊦

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 in-dom-diff (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.