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

⊦ rely(a, rs, f, g, rs0, f0, g0) ∧ rely(a, rs0, f0, g0, rs1, f1, g1) → rely(a, rs, f, g, rs1, f1, g1)

System applied rule normalize

And obtained the following new goals

Proof step 2:
rely(a, rs, f, g, rs0, f0, g0), rely(a, rs0, f0, g0, rs1, f1, g1), ¬ rely(a, rs, f, g, rs1, f1, g1) ⊦

Rule argument

Additional information

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

rely(a, rs, f, g, rs0, f0, g0), rely(a, rs0, f0, g0, rs1, f1, g1), ¬ rely(a, rs, f, g, rs1, f1, g1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (g(r) = a ∧ r ∈ rs ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f1(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-trans (previous step) (back to tree)

∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (g(r) = a ∧ r ∈ rs ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f1(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, 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)), ∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)) ⊦ (g(r) = a ∧ r ∈ rs ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f1(r))

Rule argument

pos: left 2 terms: r

Additional information

Used terms: r

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

g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, 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)), ∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)) ⊦ (g(r) = a ∧ r ∈ rs ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f1(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, g0(r) = a ∧ r ∈ rs0 → f(r) = f0(r), ∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f(r) = f1(r))

Rule argument

Additional information

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

g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, g0(r) = a ∧ r ∈ rs0 → f(r) = f0(r), ∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f(r) = f1(r))

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: r

Additional information

Used terms: r

Proof Tree for rely-trans (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.