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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  (st | s) -- a ≠ st | (s -- a) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  (st \ (dom st \ s)) -- a ≠ st \ (dom st \ (s -- a)) ⊦

Rule argument

Additional information

Used simplifier rules: restrict: ⊦ st | s = st \ (dom st \ s);

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ (a ∈ st1 → st1(a) = st2(a)))

• Proof step 5:

  ∃ a0. ¬ ( (a0 ∈ (st \ (dom st \ s)) -- a ↔ a0 ∈ st \ (dom st \ (s -- a))) ∧ (a0 ∈ (st \ (dom st \ s)) -- a → ((st \ (dom st \ s)) -- a)(a0) = (st \ (dom st \ (s -- a)))(a0))) ⊦

Rule argument

Additional information

Used terms: (st \ (dom st \ s)) -- a, st \ (dom st \ (s -- a))

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  # s ≠ 0, (st \ (dom st \ s))(a0) ≠ (st \ (dom st \ (s -- a)))(a0), a0 ≠ a, s ≠ ∅, a0 ∈ s, a0 ∈ st ⊦

Rule argument

Additional information

Used simplifier rules: Diff-ext: ⊦ a ∈ st \ s ↔ a ∈ st ∧ ¬ a ∈ s; Dom: ⊦ a ∈ dom st ↔ a ∈ st; size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic ⊦ a ≠ b → (st -- b)(a) = st(a); (s) ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; (s) ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (s) ⊦ a ∈ s -- a0 ↔ a ≠ a0 ∧ a ∈ s; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ ¬ a ∈ s → (st \ s)(a) = st(a)

• Proof step 8:

  # s ≠ 0, st(a0) ≠ (st \ (dom st \ (s -- a)))(a0), a0 ≠ a, s ≠ ∅, a0 ∈ st, a0 ∈ s, ¬ a0 ∈ dom st \ s ⊦

Rule argument

Additional information

Used simplifier rules: Diff-mems-get: ⊦ a ∈ s → (st \ s)(a) = ∅(a); Dom: ⊦ a ∈ dom st ↔ a ∈ st; ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; (s) Used terms: st, dom st \ s, a0

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

Lemma Diff-get

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  # s ≠ 0, st(a0) ≠ (st \ (dom st \ (s -- a)))(a0), a0 ≠ a, s ≠ ∅, a0 ∈ s, a0 ∈ st ⊦

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

  ⊦ ¬ a ∈ s → (st \ s)(a) = st(a)

Rule argument

Additional information

Used simplifier rules: Diff-mems-get: ⊦ a ∈ s → (st \ s)(a) = ∅(a); Dom: ⊦ a ∈ dom st ↔ a ∈ st; ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; (s) ⊦ a ∈ s -- a0 ↔ a ≠ a0 ∧ a ∈ s; (s) Used terms: st, dom st \ (s -- a), a0

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

Lemma Diff-get