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Proof step 14 from proof aux-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], a0[r => o](mk(rs1, f, g)) ⊦ (((r => o))(mk(rs1, f)) ∧ (∀ r0. r0 ∈ rs1 → g(r; a1)(r0) = a1)) ∧ R[rs0, f, g(r; a1)], (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
rs1 ∩ rs0 = ∅, a0[r => o](mk(rs1, f, g)), R[rs0, f, g] ⊦ (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0, ((r => o))(mk(rs1, f)) ∧ (r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1) ∧ R[rs0, f, g(r; a1)]

Proof step 14 from proof aux-known (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information