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Proof step 10 from proof map-in (previous step)

∃ j. (j = i0 ∨ j ∈ ints0) ∧ intfun(j) = i, i ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = i), ∀ i, intfun. i ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = i) ⊦ i = intfun(i0) ∨ i ∈ mapFun(intfun, ints0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
j ∈ ints0, ∀ i, intfun. i ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = i) ⊦ intfun(j) = intfun(i0), intfun(j) ∈ mapFun(intfun, ints0), ∃ j0. j0 ∈ ints0 ∧ intfun(j0) = intfun(j)

Proof step 10 from proof map-in (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information