Proof step 1 from proof agree-rm-01 (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(((h ≡ (h0 -- r)))(rs) ↔ ((h ≡ h0))(rs)), ¬ r ∈ rs ⊦

Rule argument

Additional information

Proof step 2 from proof agree-rm-01 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

  ⊦ ((h0 ≡ h1))(rs) ↔ (∀ r. r ∈ rs → (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0[r] = h1[r]))

• Proof step 4:

  ¬ r ∈ rs ⊦ (∀ r0. r0 ∈ rs → (r0 ∈ h ↔ r0 ∈ h0 -- r) ∧ (r0 ∈ h → h[r0] = (h0 -- r)[r0])) ↔ ((h ≡ h0))(rs)

Rule argument

Additional information

Used terms: h, h0 -- r, rs

Proof step 3 from proof agree-rm-01 (previous step) (back to tree)

Lemma agree

Proof step 4 from proof agree-rm-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  ¬ r ∈ rs ⊦ (∀ r0. r0 ∈ rs → (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = (h0 -- r)[r0])) ↔ ((h ≡ h0))(rs)

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ h -- r0 ↔ r ≠ r0 ∧ r ∈ h; (s)

Proof step 5 from proof agree-rm-01 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  ⊦ ((h0 ≡ h1))(rs) ↔ (∀ r. r ∈ rs → (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0[r] = h1[r]))

• Proof step 7:

  ¬ r ∈ rs ⊦ (∀ r0. r0 ∈ rs → (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = (h0 -- r)[r0])) ↔ (∀ r. r ∈ rs → (r ∈ h ↔ r ∈ h0) ∧ (r ∈ h → h[r] = h0[r]))

Rule argument

Additional information

Used terms: h, h0, rs

Proof step 6 from proof agree-rm-01 (previous step) (back to tree)

Lemma agree

Proof step 7 from proof agree-rm-01 (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 8:

  ¬ r ∈ rs, ∀ r0. r0 ∈ rs → (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = (h0 -- r)[r0]) ⊦ ∀ r. r ∈ rs → (r ∈ h ↔ r ∈ h0) ∧ (r ∈ h → h[r] = h0[r])

• Proof step 13:

  ¬ r ∈ rs, ∀ r. r ∈ rs → (r ∈ h ↔ r ∈ h0) ∧ (r ∈ h → h[r] = h0[r]) ⊦ ∀ r0. r0 ∈ rs → (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = (h0 -- r)[r0])

Rule argument

Additional information

Proof step 8 from proof agree-rm-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  # rs ≠ 0, rs ≠ ∅, r0 ∈ rs, ¬ r ∈ rs, ∀ r0. r0 ∈ rs → (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = (h0 -- r)[r0]) ⊦ (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = h0[r0])

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic[refset] ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic[refset]

Proof step 9 from proof agree-rm-01 (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 10:

  # rs ≠ 0, rs ≠ ∅, r0 ∈ rs, ¬ r ∈ rs, r0 ∈ rs → (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = (h0 -- r)[r0]) ⊦ (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = h0[r0])

Rule argument

Additional information

Used terms: r0

Proof step 10 from proof agree-rm-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  h[r0] = (h0 -- r)[r0], # rs ≠ 0, h[r0] ≠ h0[r0], rs ≠ ∅, r0 ∈ rs, r0 ∈ h, r0 ∈ h0, ¬ r ∈ rs ⊦

Rule argument

Additional information

Proof step 11 from proof agree-rm-01 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 12:

  r ∈ h ⊦ r ≠ r0 → (h -- r0)[r] = h[r]

Rule argument

Additional information

Used terms: h0, r, r0

Proof step 12 from proof agree-rm-01 (previous step) (back to tree)

A spec-lemma

Proof step 13 from proof agree-rm-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  # rs ≠ 0, rs ≠ ∅, r0 ∈ rs, ¬ r ∈ rs, ∀ r. r ∈ rs → (r ∈ h ↔ r ∈ h0) ∧ (r ∈ h → h[r] = h0[r]) ⊦ (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = (h0 -- r)[r0])

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic[refset] ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic[refset]

Proof step 14 from proof agree-rm-01 (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 15:

  # rs ≠ 0, rs ≠ ∅, r0 ∈ rs, ¬ r ∈ rs, r0 ∈ rs → (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = h0[r0]) ⊦ (r0 ∈ h ↔ r0 ∈ h0) ∧ (r0 ∈ h → h[r0] = (h0 -- r)[r0])

Rule argument

Additional information

Used terms: r0

Proof step 15 from proof agree-rm-01 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  h[r0] = h0[r0], # rs ≠ 0, h0[r0] ≠ (h0 -- r)[r0], rs ≠ ∅, r0 ∈ rs, r0 ∈ h, r0 ∈ h0, ¬ r ∈ rs ⊦

Rule argument

Additional information

Proof step 16 from proof agree-rm-01 (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

  r ∈ h ⊦ r ≠ r0 → (h -- r0)[r] = h[r]

Rule argument

Additional information

Used terms: h0, r, r0

Proof step 17 from proof agree-rm-01 (previous step) (back to tree)

A spec-lemma