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Proof step 7 from proof aux-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], a0[r => o][rs1, f, g] ⊦ (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0, a1[r => o][rs1, f, g(r; a1)] ∧ R[rs0, f, g(r; a1)] ∧ rs1 ∩ rs0 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
rs1 ∩ rs0 = ∅, a0[r => o][rs1, f, g], R[rs0, f, g] ⊦ (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0, a1[r => o][rs1, f, g(r; a1)] ∧ R[rs0, f, g(r; a1)]

Proof step 7 from proof aux-known (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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