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Proof step 31 from proof mapex-new (previous step)

P(H0), ∀ r1. (r1 = r ∨ r1 ∈ H ↔ r1 ∈ (∅ ++ r)[r, o] ∪ H0) ∧ (r1 = r ∨ r1 ∈ H → (H ++ r)[r1] = ((∅ ++ r)[r, o] ∪ H0)[r1]) ⊦ r = null, P(H), r ∈ H, r ∈ H0, (r0 ∈ H0 ↔ r0 ∈ H) ∧ (r0 ∈ H0 → H0[r0] = H[r0])

User applied rule all left

And obtained the following new goals

Proof step 32:
r0 = r ∨ r0 ∈ H ↔ r0 ∈ (∅ ++ r)[r, o] ∪ H0, P(H0), r0 = r ∨ r0 ∈ H → (H ++ r)[r0] = ((∅ ++ r)[r, o] ∪ H0)[r0] ⊦ r = null, P(H), r ∈ H0, r ∈ H, (r0 ∈ H0 ↔ r0 ∈ H) ∧ (r0 ∈ H0 → H0[r0] = H[r0])

Rule argument

pos: left 2 terms: r0 abort: F discard: T computed: T

Additional information

Used terms: r0