Theorem pm2.61da3ne 2597

Description: Deduction eliminating three inequalities in an antecedent. (Contributed by NM, 15-Jun-2013.)

Hypotheses

Ref	Expression
pm2.61da3ne.1	⊢ ((φ ∧ A = B) → ψ)
pm2.61da3ne.2	⊢ ((φ ∧ C = D) → ψ)
pm2.61da3ne.3	⊢ ((φ ∧ E = F) → ψ)
pm2.61da3ne.4	⊢ ((φ ∧ (A ≠ B ∧ C ≠ D ∧ E ≠ F)) → ψ)

Assertion

Ref	Expression
pm2.61da3ne	⊢ (φ → ψ)

Proof of Theorem pm2.61da3ne

Step	Hyp	Ref	Expression
1		pm2.61da3ne.1	. 2 ⊢ ((φ ∧ A = B) → ψ)
2		pm2.61da3ne.2	. 2 ⊢ ((φ ∧ C = D) → ψ)
3		pm2.61da3ne.3	. . . 4 ⊢ ((φ ∧ E = F) → ψ)
4	3	adantlr 695	. . 3 ⊢ (((φ ∧ (A ≠ B ∧ C ≠ D)) ∧ E = F) → ψ)
5		simpll 730	. . . 4 ⊢ (((φ ∧ (A ≠ B ∧ C ≠ D)) ∧ E ≠ F) → φ)
6		simplrl 736	. . . 4 ⊢ (((φ ∧ (A ≠ B ∧ C ≠ D)) ∧ E ≠ F) → A ≠ B)
7		simplrr 737	. . . 4 ⊢ (((φ ∧ (A ≠ B ∧ C ≠ D)) ∧ E ≠ F) → C ≠ D)
8		simpr 447	. . . 4 ⊢ (((φ ∧ (A ≠ B ∧ C ≠ D)) ∧ E ≠ F) → E ≠ F)
9		pm2.61da3ne.4	. . . 4 ⊢ ((φ ∧ (A ≠ B ∧ C ≠ D ∧ E ≠ F)) → ψ)
10	5, 6, 7, 8, 9	syl13anc 1184	. . 3 ⊢ (((φ ∧ (A ≠ B ∧ C ≠ D)) ∧ E ≠ F) → ψ)
11	4, 10	pm2.61dane 2595	. 2 ⊢ ((φ ∧ (A ≠ B ∧ C ≠ D)) → ψ)
12	1, 2, 11	pm2.61da2ne 2596	1 ⊢ (φ → ψ)

This proof depends on syntax axioms:   → wi 4   ∧ wa 358   ∧ w3a 934   = wceq 1642   ≠ wne 2517

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-an 360  df-3an 936  df-ne 2519

This theorem is used by: (None)