Proof step 1 from proof rely-owns-one (conclusion) (back to tree)

⊦ rely(a, rs, f, g, rs0, f0, g0) → (owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0))

System applied rule normalize

And obtained the following new goals

Proof step 2:
¬(owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)), rely(a, rs, f, g, rs0, f0, g0) ⊦

Rule argument

Additional information

Proof step 2 from proof rely-owns-one (previous step) (back to tree)

¬(owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)), rely(a, rs, f, g, rs0, f0, g0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
¬(owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

Additional information

Used simplifier rules: rely: ⊦ rely(a, rs0, f0, g0, rs1, f1, g1) ↔ (∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)));

Proof step 3 from proof rely-owns-one (previous step) (back to tree)

¬(owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 4:
g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, ¬(owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)), g(r) = a ∧ r ∈ rs → f(r) = f0(r), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

pos: left 2 terms: r

Additional information

Used terms: r

Proof step 4 from proof rely-owns-one (previous step) (back to tree)

g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, ¬(owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)), g(r) = a ∧ r ∈ rs → f(r) = f0(r), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, ¬(owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)), g0(r) = a ∧ r ∈ rs0 → f(r) = f0(r), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

Additional information

Proof step 5 from proof rely-owns-one (previous step) (back to tree)

g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, ¬(owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)), g0(r) = a ∧ r ∈ rs0 → f(r) = f0(r), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 6:
g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, owns-one(a, r, rs, g), ¬ owns-one(a, r, rs0, g0), g0(r) = a ∧ r ∈ rs0 → f(r) = f0(r), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦
Proof step 14:
g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, owns-one(a, r, rs0, g0), ¬ owns-one(a, r, rs, g), g0(r) = a ∧ r ∈ rs0 → f(r) = f0(r), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

pos: left 2

Additional information

Proof step 6 from proof rely-owns-one (previous step) (back to tree)

g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, owns-one(a, r, rs, g), ¬ owns-one(a, r, rs0, g0), g0(r) = a ∧ r ∈ rs0 → f(r) = f0(r), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r0 ∈ rs0, r ∈ rs0, r ∈ rs, ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦

Rule argument

Additional information

Used simplifier rules: owns-one: ⊦ owns-one(a, r, rs, g) ↔ r ≠ null ∧ r ∈ rs ∧ g(r) = a ∧ (∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ a);

Proof step 7 from proof rely-owns-one (previous step) (back to tree)

f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r0 ∈ rs0, r ∈ rs0, r ∈ rs, ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 8:
f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r ∈ rs, r ∈ rs0, r0 ∈ rs0, r ∈ rs ∧ r ≠ r → g(r) ≠ g(r), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r), ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

pos: left 10 terms: r

Additional information

Used terms: r

Proof step 8 from proof rely-owns-one (previous step) (back to tree)

f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r ∈ rs, r ∈ rs0, r0 ∈ rs0, r ∈ rs ∧ r ≠ r → g(r) ≠ g(r), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r), ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r0 ∈ rs0, r ∈ rs0, r ∈ rs, ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦

Rule argument

Additional information

Proof step 9 from proof rely-owns-one (previous step) (back to tree)

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 10:
g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0, f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r ∈ rs, r ∈ rs0, r0 ∈ rs0, g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r), ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

pos: left 9 terms: r

Additional information

Used terms: r

Proof step 10 from proof rely-owns-one (previous step) (back to tree)

g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0, f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r ∈ rs, r ∈ rs0, r0 ∈ rs0, g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r), ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r0 ∈ rs0, r ∈ rs0, r ∈ rs, ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦

Rule argument

Additional information

Proof step 11 from proof rely-owns-one (previous step) (back to tree)

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 12:
f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r ∈ rs, r ∈ rs0, r0 ∈ rs0, r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r), ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

pos: left 10 terms: r0

Additional information

Used terms: r0

Proof step 12 from proof rely-owns-one (previous step) (back to tree)

f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r ∈ rs, r ∈ rs0, r0 ∈ rs0, r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r), ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r0 ∈ rs0, r ∈ rs0, r ∈ rs, r0 ∈ rs → g(r0) ≠ g0(r0), ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦

Rule argument

Additional information

Proof step 13 from proof rely-owns-one (previous step) (back to tree)

f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r ≠ null, r0 ≠ r, r0 ∈ rs0, r ∈ rs0, r ∈ rs, r0 ∈ rs → g(r0) ≠ g0(r0), ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: r0

Additional information

Used terms: r0

Proof step 14 from proof rely-owns-one (previous step) (back to tree)

g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, owns-one(a, r, rs0, g0), ¬ owns-one(a, r, rs, g), g0(r) = a ∧ r ∈ rs0 → f(r) = f0(r), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r0 ∈ rs, r ∈ rs, r ∈ rs0, ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r) ⊦

Rule argument

Additional information

Used simplifier rules: owns-one: ⊦ owns-one(a, r, rs, g) ↔ r ≠ null ∧ r ∈ rs ∧ g(r) = a ∧ (∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ a);

Proof step 15 from proof rely-owns-one (previous step) (back to tree)

f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r0 ∈ rs, r ∈ rs, r ∈ rs0, ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r) ⊦

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 16:
f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r ∈ rs0, r ∈ rs, r0 ∈ rs, r ∈ rs0 ∧ r ≠ r → g0(r) ≠ g0(r), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r), ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

pos: left 10 terms: r

Additional information

Used terms: r

Proof step 16 from proof rely-owns-one (previous step) (back to tree)

f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r ∈ rs0, r ∈ rs, r0 ∈ rs, r ∈ rs0 ∧ r ≠ r → g0(r) ≠ g0(r), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r), ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r0 ∈ rs, r ∈ rs, r ∈ rs0, ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r) ⊦

Rule argument

Additional information

Proof step 17 from proof rely-owns-one (previous step) (back to tree)

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 18:
g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0, f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r ∈ rs0, r ∈ rs, r0 ∈ rs, g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r), ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

pos: left 9 terms: r

Additional information

Used terms: r

Proof step 18 from proof rely-owns-one (previous step) (back to tree)

g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0, f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r ∈ rs0, r ∈ rs, r0 ∈ rs, g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r), ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r0 ∈ rs, r ∈ rs, r ∈ rs0, ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r) ⊦

Rule argument

Additional information

Proof step 19 from proof rely-owns-one (previous step) (back to tree)

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 20:
f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r ∈ rs0, r ∈ rs, r0 ∈ rs, r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r), ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Rule argument

pos: left 10 terms: r0

Additional information

Used terms: r0

Proof step 20 from proof rely-owns-one (previous step) (back to tree)

f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r ∈ rs0, r ∈ rs, r0 ∈ rs, r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r), ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r0 ∈ rs, r ∈ rs, r ∈ rs0, r0 ∈ rs0 → g0(r0) ≠ g(r0), ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r) ⊦

Rule argument

Additional information

Proof step 21 from proof rely-owns-one (previous step) (back to tree)

f(r) = f0(r), g(r) = g(r0), g0(r) = g(r0), r ≠ null, r0 ≠ r, r0 ∈ rs, r ∈ rs, r ∈ rs0, r0 ∈ rs0 → g0(r0) ≠ g(r0), ∀ r. (g(r) = g(r0) ∧ r ∈ rs ↔ g0(r) = g(r0) ∧ r ∈ rs0) ∧ (g(r) = g(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs0 ∧ r0 ≠ r → g0(r0) ≠ g0(r) ⊦

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: r0

Additional information

Used terms: r0

Proof Tree for rely-owns-one (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.