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Proof step 41 from proof mapex-new (previous step)

r0 ∈ H ↔ r0 ∈ H0, (H ++ r)[r0] = (r0 = r ⊃ o;H0[r0]), P(H0), r0 ∈ (∅ ++ r)[r, o] ∪ H0 ⊦ r = null, P(H), r ∈ H, r ∈ H0, (r0 ∈ H0 ↔ r0 ∈ H) ∧ (r0 ∈ H0 → H0[r0] = H[r0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
(H ++ r)[r0] = H0[r0], P(H0), r0 ∈ (∅ ++ r)[r, o] ∪ H0, r0 ∈ H0, r0 ∈ H ⊦ H0[r0] = H[r0], r = null, P(H), r ∈ H0, r ∈ H

Proof step 41 from proof mapex-new (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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