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Proof step 19 from proof mod-other-both (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 20:
rs1 ∩ rs0 = ∅, P(mk(rs1, f)), R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a ⊦ g(r) = a, P(mk(rs1, f(r; o))), (a[P] * true)[rs1 ∪ rs0, f(r; o), g(r; a0)]
Proof step 27:
rs1 ∩ rs0 = ∅, P(mk(rs1, f)), R[rs0, f, g], r0 ∈ rs1, ∀ r. r ∈ rs1 → g(r) = a ⊦ (r = r0 ⊃ a0;g(r0)) = a, g(r) = a, (a[P] * true)[rs1 ∪ rs0, f(r; o), g(r; a0)]

Rule argument

pos: right 3

Proof step 19 from proof mod-other-both (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information